Mixing
A power series $sum_{n = 0}^infty a_nx^n$ such that $sum_{n=0}^infty a_n= +infty$ but $lim_{x to 1} sum_{n =...
$begingroup$
Let's consider the power series $sum_{n = 0}^{infty} a_nx^n $ with radius of convergence $1$. Moreover let's suppose that : $sum_{n = 0}^{infty} a_n= +infty$. Then I would like to find a sequence $(a_n)_{n in mathbb{N}} subseteq mathbb{R}^{mathbb{N}}$ that respect the above condition and such that :
$$lim_{x to 1, x < 1} sum_{n = 0}^infty a_nx^n ne +infty$$
First I've noticed that $a_n$ can't be a positive sequence, since if it was the case we would have for all $N$ :
$$lim_{x to 1} sum_{n = 0}^infty a_nx^n geq sum_{n = 0}^N a_n$$
Hence we need some of the $a_n$ to be negative.
Moreover I need to use the assumption that the sum at $x = 1$ diverges, because if the sum at $x = 1$ converges then Abel's theorem says that the limit at $x to 1$ and the sum of the power series at $x = 1$ are equal.
Thank you.
real-analysis calculus summation power-series
edited Jan 24 at 11:25
Somos
14.5k11336
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
add a comment |
$begingroup$
Let's consider the power series $sum_{n = 0}^{infty} a_nx^n $ with radius of convergence $1$. Moreover let's suppose that : $sum_{n = 0}^{infty} a_n= +infty$. Then I would like to find a sequence $(a_n)_{n in mathbb{N}} subseteq mathbb{R}^{mathbb{N}}$ that respect the above condition and such that :
$$lim_{x to 1, x < 1} sum_{n = 0}^infty a_nx^n ne +infty$$
First I've noticed that $a_n$ can't be a positive sequence, since if it was the case we would have for all $N$ :
$$lim_{x to 1} sum_{n = 0}^infty a_nx^n geq sum_{n = 0}^N a_n$$
Hence we need some of the $a_n$ to be negative.
Moreover I need to use the assumption that the sum at $x = 1$ diverges, because if the sum at $x = 1$ converges then Abel's theorem says that the limit at $x to 1$ and the sum of the power series at $x = 1$ are equal.
Thank you.
real-analysis calculus summation power-series
edited Jan 24 at 11:25
Somos
14.5k11336
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
$begingroup$
Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round?
$endgroup$
– maxmilgram
Jan 24 at 10:04
$begingroup$
You are searching for an Abel summable sequence.
$endgroup$
– Daniele Tampieri
Jan 24 at 10:41
1
$begingroup$
@maxmilgram No, though it takes a moment to realize that. You have $lim_{xto 1}sum_{n=0}^N a_nx^n=sum_{n=0}^Na_n$ (just polynomials). This means $sum_{n=0}^N a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for some $delta$, corresponding to a chose $epsilon > 0$. With all $a_n$ positive, we get $sum_{n=0}^infty a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for that $delta$. From that follows what's in the post, because $epsilon > 0$ can be chosen freely.
$endgroup$
– Ingix
Jan 24 at 12:25
$begingroup$
Writing $$lim_{x to 1} sum_{n = 0}^infty a_nx^n ne infty$$ is a little strange. Why not ask: Does $$lim_{x to 1^-} sum_{n = 0}^infty a_nx^n =infty?$$
$endgroup$
– zhw.
Jan 24 at 21:52
add a comment |
$begingroup$
Let's consider the power series $sum_{n = 0}^{infty} a_nx^n $ with radius of convergence $1$. Moreover let's suppose that : $sum_{n = 0}^{infty} a_n= +infty$. Then I would like to find a sequence $(a_n)_{n in mathbb{N}} subseteq mathbb{R}^{mathbb{N}}$ that respect the above condition and such that :
$$lim_{x to 1, x < 1} sum_{n = 0}^infty a_nx^n ne +infty$$
First I've noticed that $a_n$ can't be a positive sequence, since if it was the case we would have for all $N$ :
$$lim_{x to 1} sum_{n = 0}^infty a_nx^n geq sum_{n = 0}^N a_n$$
Hence we need some of the $a_n$ to be negative.
Moreover I need to use the assumption that the sum at $x = 1$ diverges, because if the sum at $x = 1$ converges then Abel's theorem says that the limit at $x to 1$ and the sum of the power series at $x = 1$ are equal.
Thank you.
real-analysis calculus summation power-series
edited Jan 24 at 11:25
Somos
14.5k11336
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
$endgroup$
Let's consider the power series $sum_{n = 0}^{infty} a_nx^n $ with radius of convergence $1$. Moreover let's suppose that : $sum_{n = 0}^{infty} a_n= +infty$. Then I would like to find a sequence $(a_n)_{n in mathbb{N}} subseteq mathbb{R}^{mathbb{N}}$ that respect the above condition and such that :
$$lim_{x to 1, x < 1} sum_{n = 0}^infty a_nx^n ne +infty$$
First I've noticed that $a_n$ can't be a positive sequence, since if it was the case we would have for all $N$ :
$$lim_{x to 1} sum_{n = 0}^infty a_nx^n geq sum_{n = 0}^N a_n$$
Hence we need some of the $a_n$ to be negative.
Moreover I need to use the assumption that the sum at $x = 1$ diverges, because if the sum at $x = 1$ converges then Abel's theorem says that the limit at $x to 1$ and the sum of the power series at $x = 1$ are equal.
Thank you.
real-analysis calculus summation power-series
real-analysis calculus summation power-series
edited Jan 24 at 11:25
Somos
14.5k11336
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
edited Jan 24 at 11:25
Somos
14.5k11336
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
edited Jan 24 at 11:25
Somos
14.5k11336
edited Jan 24 at 11:25
Somos
14.5k11336
edited Jan 24 at 11:25
Somos
14.5k11336
14.5k11336
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
asked Jan 24 at 8:41
dghkgfzyukzdghkgfzyukz
16112
16112
$begingroup$
Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round?
$endgroup$
– maxmilgram
Jan 24 at 10:04
$begingroup$
You are searching for an Abel summable sequence.
$endgroup$
– Daniele Tampieri
Jan 24 at 10:41
1
$begingroup$
@maxmilgram No, though it takes a moment to realize that. You have $lim_{xto 1}sum_{n=0}^N a_nx^n=sum_{n=0}^Na_n$ (just polynomials). This means $sum_{n=0}^N a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for some $delta$, corresponding to a chose $epsilon > 0$. With all $a_n$ positive, we get $sum_{n=0}^infty a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for that $delta$. From that follows what's in the post, because $epsilon > 0$ can be chosen freely.
$endgroup$
– Ingix
Jan 24 at 12:25
$begingroup$
Writing $$lim_{x to 1} sum_{n = 0}^infty a_nx^n ne infty$$ is a little strange. Why not ask: Does $$lim_{x to 1^-} sum_{n = 0}^infty a_nx^n =infty?$$
$endgroup$
– zhw.
Jan 24 at 21:52
add a comment |
$begingroup$
Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round?
$endgroup$
– maxmilgram
Jan 24 at 10:04
$begingroup$
You are searching for an Abel summable sequence.
$endgroup$
– Daniele Tampieri
Jan 24 at 10:41
1
$begingroup$
@maxmilgram No, though it takes a moment to realize that. You have $lim_{xto 1}sum_{n=0}^N a_nx^n=sum_{n=0}^Na_n$ (just polynomials). This means $sum_{n=0}^N a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for some $delta$, corresponding to a chose $epsilon > 0$. With all $a_n$ positive, we get $sum_{n=0}^infty a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for that $delta$. From that follows what's in the post, because $epsilon > 0$ can be chosen freely.
$endgroup$
– Ingix
Jan 24 at 12:25
$begingroup$
Writing $$lim_{x to 1} sum_{n = 0}^infty a_nx^n ne infty$$ is a little strange. Why not ask: Does $$lim_{x to 1^-} sum_{n = 0}^infty a_nx^n =infty?$$
$endgroup$
– zhw.
Jan 24 at 21:52
$begingroup$
Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round?
$endgroup$
– maxmilgram
Jan 24 at 10:04
$begingroup$
Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round?
$endgroup$
– maxmilgram
Jan 24 at 10:04
$begingroup$
You are searching for an Abel summable sequence.
$endgroup$
– Daniele Tampieri
Jan 24 at 10:41
$begingroup$
You are searching for an Abel summable sequence.
$endgroup$
– Daniele Tampieri
Jan 24 at 10:41
1
1
$begingroup$
@maxmilgram No, though it takes a moment to realize that. You have $lim_{xto 1}sum_{n=0}^N a_nx^n=sum_{n=0}^Na_n$ (just polynomials). This means $sum_{n=0}^N a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for some $delta$, corresponding to a chose $epsilon > 0$. With all $a_n$ positive, we get $sum_{n=0}^infty a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for that $delta$. From that follows what's in the post, because $epsilon > 0$ can be chosen freely.
$endgroup$
– Ingix
Jan 24 at 12:25
$begingroup$
@maxmilgram No, though it takes a moment to realize that. You have $lim_{xto 1}sum_{n=0}^N a_nx^n=sum_{n=0}^Na_n$ (just polynomials). This means $sum_{n=0}^N a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for some $delta$, corresponding to a chose $epsilon > 0$. With all $a_n$ positive, we get $sum_{n=0}^infty a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for that $delta$. From that follows what's in the post, because $epsilon > 0$ can be chosen freely.
$endgroup$
– Ingix
Jan 24 at 12:25
$begingroup$
Writing $$lim_{x to 1} sum_{n = 0}^infty a_nx^n ne infty$$ is a little strange. Why not ask: Does $$lim_{x to 1^-} sum_{n = 0}^infty a_nx^n =infty?$$
$endgroup$
– zhw.
Jan 24 at 21:52
$begingroup$
Writing $$lim_{x to 1} sum_{n = 0}^infty a_nx^n ne infty$$ is a little strange. Why not ask: Does $$lim_{x to 1^-} sum_{n = 0}^infty a_nx^n =infty?$$
$endgroup$
– zhw.
Jan 24 at 21:52
add a comment |
2 Answers
2
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oldest
votes
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Such sequences don't exist. We prove this by contradiction:
Assume that $A_n := sum_{k=1}^n a_k rightarrow infty$, $ f(x):=sum_{k=1}^infty a_k z^k$ is convergent in the unit disc and $lim_{x uparrow 1} f(x)$ exists. Removing finite many $a_i$ doesn't change the behaviour of $A_n$ and also not the existence of the limes for $x uparrow 1$. Thus, we may assume that $A_n ge 0$ for all $n ge 1$. Now use Abel summation in the form
$$label{1}tag{1}sum_{k=1}^n a_k x^k = A_n x^n - sum_{k=1}^{n-1} A_k x^k (1 - x).$$
Because
$$|A_n x^n| le sum_{j=1}^n |a_j| |x|^j$$
and the series is absolute convergent (as a power series) for $|x| < 1$, we see that $|A_n x^n|$ is bounded for fixed $|x| < 1$. Since $|A_n x^n| le C(x)$, we get for all $|y| < |x|$ that $|A_n y^n| le C(x) |y/x|^n rightarrow 0$. Hence $lim_{n rightarrow infty } A_n y^n =0$ for all $|y| < |x|$. Since $x$ was arbitary, we get this statement for all $|y| <1$ (not necessarily uniform convergence).
Letting $n rightarrow infty$ in eqref{1} gives for $|x| <1$ that
$$label{2}tag{2}f(x)=sum_{k=1}^infty a_k x^k = (1-x) sum_{k=1}^infty A_k x^k.$$
Since $lim_{x uparrow 1} f(x):=c$ exists, we have
$$sum_{k=1}^infty A_k x^k sim frac{1}{1-x}.$$
The Hardy–Littlewood tauberian theorem already implies that
$$sum_{k=1}^n (n+1-k) a_k = sum_{k=1}^n A_k sim n.$$
But, we have
$$frac{1}{2n}sum_{k=1}^{2n} A_k ge frac{1}{2} A_n rightarrow infty.$$
A contradiction!
The problem changes rapidly, if we only require that $sum_{k=1}^n a_k$ is not convergent.
For example take $a_k = (-1)^k$: We have
$$sum_{k=0}^infty (-1)^k x^k = frac{1}{x+1}$$
and that $lim_{x uparrow 1} (x+1)^{-1} = 1/2$, but $sum_{k=0}^n (-1)^k$ is not convergent.
As zhw shows in the second answer, we can also prove that $lim_{x uparrow 1} f(x) = infty$.
Note for this that the identity in eqref{2} implies, since $A_n >K$ for all $n ge N$ that
$$f(x) ge (1-x) K sum_{k=N}^infty x^k = K x^N$$
and thus $liminf_{x uparrow 1} f(x) ge K$. Because $K>0$ is arbitary large, we get already $lim_{x uparrow 1} f(x) = infty$.
answered Jan 24 at 13:54
p4schp4sch
5,440318
$endgroup$
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How can you assume there is a finite a_i that a_i<0?
$endgroup$
– Shaq
Jan 24 at 18:44
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We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n ge N$ for some $N in mathbb{N}$, because $A_n rightarrow infty$. So removing all negative numbers of $a_1,ldots,a_N$ gives $A_n ge 0$ for all $n in mathbb{N}$.
$endgroup$
– p4sch
Jan 24 at 20:06
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You are assuming $lim_{uparrow 1}f(x)$ exists?
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– zhw.
Jan 24 at 21:46
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Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution.
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– p4sch
Jan 25 at 10:18
add a comment |
$begingroup$
The following proof that $lim_{xto 1^-}sum_{n=0}^{infty}a_nx^n = infty$ seems simpler to me..
Lemma: Let $A_n=sum_{k=0}^{n}|a_k|.$ Then $sum_{n=0}^{infty}A_nx^n<infty$ for $xin (0,1).$
This follows from the fact that the radius of convergence is $1,$ which is the same as saying $limsup |a_n|^{1/n} =1.$ This is a nice exercise. (A proof of the lemma is now in the comments.)
To prove the main result, let $S_n=sum_{k=0}^{n}a_n;$ set $S_{-1}=0.$ Let $xin (0,1).$ Then
$$tag 1sum_{n=0}^{infty}a_nx^n = sum_{n=0}^{infty}(S_n-S_{n-1})x^n.$$
Now because $|S_n| le A_n,$ the lemma shows we can write the last series as the difference of two convergent series, i.e. as
$$ sum_{n=0}^{infty}S_nx^n - sum_{n=1}^{infty}S_{n-1}x^n = sum_{n=0}^{infty}S_nx^n - sum_{n=0}^{infty}S_{n}x^{n+1} = sum_{n=0}^{infty}S_nx^n(1-x).$$
Now let $M>0.$ Then there is $N$ such $S_n>M$ for $n>N.$ Write the last series as
$$(1-x)left (sum_{n=0}^{N}S_nx^n +sum_{n=N+1}^{infty}S_nx^nright ) > (1-x)left (sum_{n=0}^{N}S_nx^n +Mx^{N+1}frac{1}{1-x}right ).$$
The $liminf_{xto 1^-}$ of the expression on the right equals $0 + M =M.$ We have thus shown the $liminf_{xto 1^-}$ of the left side of $(1)$ is $ge M.$ Since $M$ was arbitrary, this $liminf$ is $infty.$ Thus the limit of left side of $(1)$ is $infty$ as desired.
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
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You mean $A_n = sum_{ k = 0}^n mid a_k mid$, no ?
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– Thinking
Jan 24 at 21:58
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@Thinking Yes, thanks. Will correct.
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– zhw.
Jan 24 at 22:00
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You didn't completely correct the typo. I know that if we denote $K_n = sum_{k = 0}^n a_k$ then $sum_{n = 0}^infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$.
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– Thinking
Jan 24 at 22:17
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@Thinking I certainly corrected the typo. It seems you have another question. I'll be back.
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– zhw.
Jan 24 at 22:22
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? So $A_n$ is just $(n+1)mid a_n mid$ ?
$endgroup$
– Thinking
Jan 24 at 22:23
|
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2 Answers
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2 Answers
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$begingroup$
Such sequences don't exist. We prove this by contradiction:
Assume that $A_n := sum_{k=1}^n a_k rightarrow infty$, $ f(x):=sum_{k=1}^infty a_k z^k$ is convergent in the unit disc and $lim_{x uparrow 1} f(x)$ exists. Removing finite many $a_i$ doesn't change the behaviour of $A_n$ and also not the existence of the limes for $x uparrow 1$. Thus, we may assume that $A_n ge 0$ for all $n ge 1$. Now use Abel summation in the form
$$label{1}tag{1}sum_{k=1}^n a_k x^k = A_n x^n - sum_{k=1}^{n-1} A_k x^k (1 - x).$$
Because
$$|A_n x^n| le sum_{j=1}^n |a_j| |x|^j$$
and the series is absolute convergent (as a power series) for $|x| < 1$, we see that $|A_n x^n|$ is bounded for fixed $|x| < 1$. Since $|A_n x^n| le C(x)$, we get for all $|y| < |x|$ that $|A_n y^n| le C(x) |y/x|^n rightarrow 0$. Hence $lim_{n rightarrow infty } A_n y^n =0$ for all $|y| < |x|$. Since $x$ was arbitary, we get this statement for all $|y| <1$ (not necessarily uniform convergence).
Letting $n rightarrow infty$ in eqref{1} gives for $|x| <1$ that
$$label{2}tag{2}f(x)=sum_{k=1}^infty a_k x^k = (1-x) sum_{k=1}^infty A_k x^k.$$
Since $lim_{x uparrow 1} f(x):=c$ exists, we have
$$sum_{k=1}^infty A_k x^k sim frac{1}{1-x}.$$
The Hardy–Littlewood tauberian theorem already implies that
$$sum_{k=1}^n (n+1-k) a_k = sum_{k=1}^n A_k sim n.$$
But, we have
$$frac{1}{2n}sum_{k=1}^{2n} A_k ge frac{1}{2} A_n rightarrow infty.$$
A contradiction!
The problem changes rapidly, if we only require that $sum_{k=1}^n a_k$ is not convergent.
For example take $a_k = (-1)^k$: We have
$$sum_{k=0}^infty (-1)^k x^k = frac{1}{x+1}$$
and that $lim_{x uparrow 1} (x+1)^{-1} = 1/2$, but $sum_{k=0}^n (-1)^k$ is not convergent.
As zhw shows in the second answer, we can also prove that $lim_{x uparrow 1} f(x) = infty$.
Note for this that the identity in eqref{2} implies, since $A_n >K$ for all $n ge N$ that
$$f(x) ge (1-x) K sum_{k=N}^infty x^k = K x^N$$
and thus $liminf_{x uparrow 1} f(x) ge K$. Because $K>0$ is arbitary large, we get already $lim_{x uparrow 1} f(x) = infty$.
answered Jan 24 at 13:54
p4schp4sch
5,440318
$endgroup$
$begingroup$
How can you assume there is a finite a_i that a_i<0?
$endgroup$
– Shaq
Jan 24 at 18:44
$begingroup$
We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n ge N$ for some $N in mathbb{N}$, because $A_n rightarrow infty$. So removing all negative numbers of $a_1,ldots,a_N$ gives $A_n ge 0$ for all $n in mathbb{N}$.
$endgroup$
– p4sch
Jan 24 at 20:06
$begingroup$
You are assuming $lim_{uparrow 1}f(x)$ exists?
$endgroup$
– zhw.
Jan 24 at 21:46
$begingroup$
Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution.
$endgroup$
– p4sch
Jan 25 at 10:18
add a comment |
$begingroup$
Such sequences don't exist. We prove this by contradiction:
Assume that $A_n := sum_{k=1}^n a_k rightarrow infty$, $ f(x):=sum_{k=1}^infty a_k z^k$ is convergent in the unit disc and $lim_{x uparrow 1} f(x)$ exists. Removing finite many $a_i$ doesn't change the behaviour of $A_n$ and also not the existence of the limes for $x uparrow 1$. Thus, we may assume that $A_n ge 0$ for all $n ge 1$. Now use Abel summation in the form
$$label{1}tag{1}sum_{k=1}^n a_k x^k = A_n x^n - sum_{k=1}^{n-1} A_k x^k (1 - x).$$
Because
$$|A_n x^n| le sum_{j=1}^n |a_j| |x|^j$$
and the series is absolute convergent (as a power series) for $|x| < 1$, we see that $|A_n x^n|$ is bounded for fixed $|x| < 1$. Since $|A_n x^n| le C(x)$, we get for all $|y| < |x|$ that $|A_n y^n| le C(x) |y/x|^n rightarrow 0$. Hence $lim_{n rightarrow infty } A_n y^n =0$ for all $|y| < |x|$. Since $x$ was arbitary, we get this statement for all $|y| <1$ (not necessarily uniform convergence).
Letting $n rightarrow infty$ in eqref{1} gives for $|x| <1$ that
$$label{2}tag{2}f(x)=sum_{k=1}^infty a_k x^k = (1-x) sum_{k=1}^infty A_k x^k.$$
Since $lim_{x uparrow 1} f(x):=c$ exists, we have
$$sum_{k=1}^infty A_k x^k sim frac{1}{1-x}.$$
The Hardy–Littlewood tauberian theorem already implies that
$$sum_{k=1}^n (n+1-k) a_k = sum_{k=1}^n A_k sim n.$$
But, we have
$$frac{1}{2n}sum_{k=1}^{2n} A_k ge frac{1}{2} A_n rightarrow infty.$$
A contradiction!
The problem changes rapidly, if we only require that $sum_{k=1}^n a_k$ is not convergent.
For example take $a_k = (-1)^k$: We have
$$sum_{k=0}^infty (-1)^k x^k = frac{1}{x+1}$$
and that $lim_{x uparrow 1} (x+1)^{-1} = 1/2$, but $sum_{k=0}^n (-1)^k$ is not convergent.
As zhw shows in the second answer, we can also prove that $lim_{x uparrow 1} f(x) = infty$.
Note for this that the identity in eqref{2} implies, since $A_n >K$ for all $n ge N$ that
$$f(x) ge (1-x) K sum_{k=N}^infty x^k = K x^N$$
and thus $liminf_{x uparrow 1} f(x) ge K$. Because $K>0$ is arbitary large, we get already $lim_{x uparrow 1} f(x) = infty$.
answered Jan 24 at 13:54
p4schp4sch
5,440318
$endgroup$
$begingroup$
How can you assume there is a finite a_i that a_i<0?
$endgroup$
– Shaq
Jan 24 at 18:44
$begingroup$
We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n ge N$ for some $N in mathbb{N}$, because $A_n rightarrow infty$. So removing all negative numbers of $a_1,ldots,a_N$ gives $A_n ge 0$ for all $n in mathbb{N}$.
$endgroup$
– p4sch
Jan 24 at 20:06
$begingroup$
You are assuming $lim_{uparrow 1}f(x)$ exists?
$endgroup$
– zhw.
Jan 24 at 21:46
$begingroup$
Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution.
$endgroup$
– p4sch
Jan 25 at 10:18
add a comment |
$begingroup$
Such sequences don't exist. We prove this by contradiction:
Assume that $A_n := sum_{k=1}^n a_k rightarrow infty$, $ f(x):=sum_{k=1}^infty a_k z^k$ is convergent in the unit disc and $lim_{x uparrow 1} f(x)$ exists. Removing finite many $a_i$ doesn't change the behaviour of $A_n$ and also not the existence of the limes for $x uparrow 1$. Thus, we may assume that $A_n ge 0$ for all $n ge 1$. Now use Abel summation in the form
$$label{1}tag{1}sum_{k=1}^n a_k x^k = A_n x^n - sum_{k=1}^{n-1} A_k x^k (1 - x).$$
Because
$$|A_n x^n| le sum_{j=1}^n |a_j| |x|^j$$
and the series is absolute convergent (as a power series) for $|x| < 1$, we see that $|A_n x^n|$ is bounded for fixed $|x| < 1$. Since $|A_n x^n| le C(x)$, we get for all $|y| < |x|$ that $|A_n y^n| le C(x) |y/x|^n rightarrow 0$. Hence $lim_{n rightarrow infty } A_n y^n =0$ for all $|y| < |x|$. Since $x$ was arbitary, we get this statement for all $|y| <1$ (not necessarily uniform convergence).
Letting $n rightarrow infty$ in eqref{1} gives for $|x| <1$ that
$$label{2}tag{2}f(x)=sum_{k=1}^infty a_k x^k = (1-x) sum_{k=1}^infty A_k x^k.$$
Since $lim_{x uparrow 1} f(x):=c$ exists, we have
$$sum_{k=1}^infty A_k x^k sim frac{1}{1-x}.$$
The Hardy–Littlewood tauberian theorem already implies that
$$sum_{k=1}^n (n+1-k) a_k = sum_{k=1}^n A_k sim n.$$
But, we have
$$frac{1}{2n}sum_{k=1}^{2n} A_k ge frac{1}{2} A_n rightarrow infty.$$
A contradiction!
The problem changes rapidly, if we only require that $sum_{k=1}^n a_k$ is not convergent.
For example take $a_k = (-1)^k$: We have
$$sum_{k=0}^infty (-1)^k x^k = frac{1}{x+1}$$
and that $lim_{x uparrow 1} (x+1)^{-1} = 1/2$, but $sum_{k=0}^n (-1)^k$ is not convergent.
As zhw shows in the second answer, we can also prove that $lim_{x uparrow 1} f(x) = infty$.
Note for this that the identity in eqref{2} implies, since $A_n >K$ for all $n ge N$ that
$$f(x) ge (1-x) K sum_{k=N}^infty x^k = K x^N$$
and thus $liminf_{x uparrow 1} f(x) ge K$. Because $K>0$ is arbitary large, we get already $lim_{x uparrow 1} f(x) = infty$.
answered Jan 24 at 13:54
p4schp4sch
5,440318
$endgroup$
Such sequences don't exist. We prove this by contradiction:
Assume that $A_n := sum_{k=1}^n a_k rightarrow infty$, $ f(x):=sum_{k=1}^infty a_k z^k$ is convergent in the unit disc and $lim_{x uparrow 1} f(x)$ exists. Removing finite many $a_i$ doesn't change the behaviour of $A_n$ and also not the existence of the limes for $x uparrow 1$. Thus, we may assume that $A_n ge 0$ for all $n ge 1$. Now use Abel summation in the form
$$label{1}tag{1}sum_{k=1}^n a_k x^k = A_n x^n - sum_{k=1}^{n-1} A_k x^k (1 - x).$$
Because
$$|A_n x^n| le sum_{j=1}^n |a_j| |x|^j$$
and the series is absolute convergent (as a power series) for $|x| < 1$, we see that $|A_n x^n|$ is bounded for fixed $|x| < 1$. Since $|A_n x^n| le C(x)$, we get for all $|y| < |x|$ that $|A_n y^n| le C(x) |y/x|^n rightarrow 0$. Hence $lim_{n rightarrow infty } A_n y^n =0$ for all $|y| < |x|$. Since $x$ was arbitary, we get this statement for all $|y| <1$ (not necessarily uniform convergence).
Letting $n rightarrow infty$ in eqref{1} gives for $|x| <1$ that
$$label{2}tag{2}f(x)=sum_{k=1}^infty a_k x^k = (1-x) sum_{k=1}^infty A_k x^k.$$
Since $lim_{x uparrow 1} f(x):=c$ exists, we have
$$sum_{k=1}^infty A_k x^k sim frac{1}{1-x}.$$
The Hardy–Littlewood tauberian theorem already implies that
$$sum_{k=1}^n (n+1-k) a_k = sum_{k=1}^n A_k sim n.$$
But, we have
$$frac{1}{2n}sum_{k=1}^{2n} A_k ge frac{1}{2} A_n rightarrow infty.$$
A contradiction!
The problem changes rapidly, if we only require that $sum_{k=1}^n a_k$ is not convergent.
For example take $a_k = (-1)^k$: We have
$$sum_{k=0}^infty (-1)^k x^k = frac{1}{x+1}$$
and that $lim_{x uparrow 1} (x+1)^{-1} = 1/2$, but $sum_{k=0}^n (-1)^k$ is not convergent.
As zhw shows in the second answer, we can also prove that $lim_{x uparrow 1} f(x) = infty$.
Note for this that the identity in eqref{2} implies, since $A_n >K$ for all $n ge N$ that
$$f(x) ge (1-x) K sum_{k=N}^infty x^k = K x^N$$
and thus $liminf_{x uparrow 1} f(x) ge K$. Because $K>0$ is arbitary large, we get already $lim_{x uparrow 1} f(x) = infty$.
answered Jan 24 at 13:54
p4schp4sch
5,440318
edited Jan 25 at 10:17
answered Jan 24 at 13:54
p4schp4sch
5,440318
answered Jan 24 at 13:54
p4schp4sch
5,440318
answered Jan 24 at 13:54
p4schp4sch
5,440318
5,440318
$begingroup$
How can you assume there is a finite a_i that a_i<0?
$endgroup$
– Shaq
Jan 24 at 18:44
$begingroup$
We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n ge N$ for some $N in mathbb{N}$, because $A_n rightarrow infty$. So removing all negative numbers of $a_1,ldots,a_N$ gives $A_n ge 0$ for all $n in mathbb{N}$.
$endgroup$
– p4sch
Jan 24 at 20:06
$begingroup$
You are assuming $lim_{uparrow 1}f(x)$ exists?
$endgroup$
– zhw.
Jan 24 at 21:46
$begingroup$
Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution.
$endgroup$
– p4sch
Jan 25 at 10:18
add a comment |
$begingroup$
How can you assume there is a finite a_i that a_i<0?
$endgroup$
– Shaq
Jan 24 at 18:44
$begingroup$
We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n ge N$ for some $N in mathbb{N}$, because $A_n rightarrow infty$. So removing all negative numbers of $a_1,ldots,a_N$ gives $A_n ge 0$ for all $n in mathbb{N}$.
$endgroup$
– p4sch
Jan 24 at 20:06
$begingroup$
You are assuming $lim_{uparrow 1}f(x)$ exists?
$endgroup$
– zhw.
Jan 24 at 21:46
$begingroup$
Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution.
$endgroup$
– p4sch
Jan 25 at 10:18
$begingroup$
How can you assume there is a finite a_i that a_i<0?
$endgroup$
– Shaq
Jan 24 at 18:44
$begingroup$
How can you assume there is a finite a_i that a_i<0?
$endgroup$
– Shaq
Jan 24 at 18:44
$begingroup$
We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n ge N$ for some $N in mathbb{N}$, because $A_n rightarrow infty$. So removing all negative numbers of $a_1,ldots,a_N$ gives $A_n ge 0$ for all $n in mathbb{N}$.
$endgroup$
– p4sch
Jan 24 at 20:06
$begingroup$
We don't know that $a_i <0$ only for finite many $i$, but we know that $A_n >0$ for all $n ge N$ for some $N in mathbb{N}$, because $A_n rightarrow infty$. So removing all negative numbers of $a_1,ldots,a_N$ gives $A_n ge 0$ for all $n in mathbb{N}$.
$endgroup$
– p4sch
Jan 24 at 20:06
$begingroup$
You are assuming $lim_{uparrow 1}f(x)$ exists?
$endgroup$
– zhw.
Jan 24 at 21:46
$begingroup$
You are assuming $lim_{uparrow 1}f(x)$ exists?
$endgroup$
– zhw.
Jan 24 at 21:46
$begingroup$
Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution.
$endgroup$
– p4sch
Jan 25 at 10:18
$begingroup$
Yes, my argument shows that the limes doesn't exist. In fact, your argument shows that the sequence tend to $+infty$. I have added also your argument to my answer. In fact, it is a more straight forward solution.
$endgroup$
– p4sch
Jan 25 at 10:18
add a comment |
$begingroup$
The following proof that $lim_{xto 1^-}sum_{n=0}^{infty}a_nx^n = infty$ seems simpler to me..
Lemma: Let $A_n=sum_{k=0}^{n}|a_k|.$ Then $sum_{n=0}^{infty}A_nx^n<infty$ for $xin (0,1).$
This follows from the fact that the radius of convergence is $1,$ which is the same as saying $limsup |a_n|^{1/n} =1.$ This is a nice exercise. (A proof of the lemma is now in the comments.)
To prove the main result, let $S_n=sum_{k=0}^{n}a_n;$ set $S_{-1}=0.$ Let $xin (0,1).$ Then
$$tag 1sum_{n=0}^{infty}a_nx^n = sum_{n=0}^{infty}(S_n-S_{n-1})x^n.$$
Now because $|S_n| le A_n,$ the lemma shows we can write the last series as the difference of two convergent series, i.e. as
$$ sum_{n=0}^{infty}S_nx^n - sum_{n=1}^{infty}S_{n-1}x^n = sum_{n=0}^{infty}S_nx^n - sum_{n=0}^{infty}S_{n}x^{n+1} = sum_{n=0}^{infty}S_nx^n(1-x).$$
Now let $M>0.$ Then there is $N$ such $S_n>M$ for $n>N.$ Write the last series as
$$(1-x)left (sum_{n=0}^{N}S_nx^n +sum_{n=N+1}^{infty}S_nx^nright ) > (1-x)left (sum_{n=0}^{N}S_nx^n +Mx^{N+1}frac{1}{1-x}right ).$$
The $liminf_{xto 1^-}$ of the expression on the right equals $0 + M =M.$ We have thus shown the $liminf_{xto 1^-}$ of the left side of $(1)$ is $ge M.$ Since $M$ was arbitrary, this $liminf$ is $infty.$ Thus the limit of left side of $(1)$ is $infty$ as desired.
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
You mean $A_n = sum_{ k = 0}^n mid a_k mid$, no ?
$endgroup$
– Thinking
Jan 24 at 21:58
$begingroup$
@Thinking Yes, thanks. Will correct.
$endgroup$
– zhw.
Jan 24 at 22:00
$begingroup$
You didn't completely correct the typo. I know that if we denote $K_n = sum_{k = 0}^n a_k$ then $sum_{n = 0}^infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$.
$endgroup$
– Thinking
Jan 24 at 22:17
$begingroup$
@Thinking I certainly corrected the typo. It seems you have another question. I'll be back.
$endgroup$
– zhw.
Jan 24 at 22:22
$begingroup$
? So $A_n$ is just $(n+1)mid a_n mid$ ?
$endgroup$
– Thinking
Jan 24 at 22:23
|
show 4 more comments
$begingroup$
The following proof that $lim_{xto 1^-}sum_{n=0}^{infty}a_nx^n = infty$ seems simpler to me..
Lemma: Let $A_n=sum_{k=0}^{n}|a_k|.$ Then $sum_{n=0}^{infty}A_nx^n<infty$ for $xin (0,1).$
This follows from the fact that the radius of convergence is $1,$ which is the same as saying $limsup |a_n|^{1/n} =1.$ This is a nice exercise. (A proof of the lemma is now in the comments.)
To prove the main result, let $S_n=sum_{k=0}^{n}a_n;$ set $S_{-1}=0.$ Let $xin (0,1).$ Then
$$tag 1sum_{n=0}^{infty}a_nx^n = sum_{n=0}^{infty}(S_n-S_{n-1})x^n.$$
Now because $|S_n| le A_n,$ the lemma shows we can write the last series as the difference of two convergent series, i.e. as
$$ sum_{n=0}^{infty}S_nx^n - sum_{n=1}^{infty}S_{n-1}x^n = sum_{n=0}^{infty}S_nx^n - sum_{n=0}^{infty}S_{n}x^{n+1} = sum_{n=0}^{infty}S_nx^n(1-x).$$
Now let $M>0.$ Then there is $N$ such $S_n>M$ for $n>N.$ Write the last series as
$$(1-x)left (sum_{n=0}^{N}S_nx^n +sum_{n=N+1}^{infty}S_nx^nright ) > (1-x)left (sum_{n=0}^{N}S_nx^n +Mx^{N+1}frac{1}{1-x}right ).$$
The $liminf_{xto 1^-}$ of the expression on the right equals $0 + M =M.$ We have thus shown the $liminf_{xto 1^-}$ of the left side of $(1)$ is $ge M.$ Since $M$ was arbitrary, this $liminf$ is $infty.$ Thus the limit of left side of $(1)$ is $infty$ as desired.
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
$endgroup$
$begingroup$
You mean $A_n = sum_{ k = 0}^n mid a_k mid$, no ?
$endgroup$
– Thinking
Jan 24 at 21:58
$begingroup$
@Thinking Yes, thanks. Will correct.
$endgroup$
– zhw.
Jan 24 at 22:00
$begingroup$
You didn't completely correct the typo. I know that if we denote $K_n = sum_{k = 0}^n a_k$ then $sum_{n = 0}^infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$.
$endgroup$
– Thinking
Jan 24 at 22:17
$begingroup$
@Thinking I certainly corrected the typo. It seems you have another question. I'll be back.
$endgroup$
– zhw.
Jan 24 at 22:22
$begingroup$
? So $A_n$ is just $(n+1)mid a_n mid$ ?
$endgroup$
– Thinking
Jan 24 at 22:23
|
show 4 more comments
$begingroup$
The following proof that $lim_{xto 1^-}sum_{n=0}^{infty}a_nx^n = infty$ seems simpler to me..
Lemma: Let $A_n=sum_{k=0}^{n}|a_k|.$ Then $sum_{n=0}^{infty}A_nx^n<infty$ for $xin (0,1).$
This follows from the fact that the radius of convergence is $1,$ which is the same as saying $limsup |a_n|^{1/n} =1.$ This is a nice exercise. (A proof of the lemma is now in the comments.)
To prove the main result, let $S_n=sum_{k=0}^{n}a_n;$ set $S_{-1}=0.$ Let $xin (0,1).$ Then
$$tag 1sum_{n=0}^{infty}a_nx^n = sum_{n=0}^{infty}(S_n-S_{n-1})x^n.$$
Now because $|S_n| le A_n,$ the lemma shows we can write the last series as the difference of two convergent series, i.e. as
$$ sum_{n=0}^{infty}S_nx^n - sum_{n=1}^{infty}S_{n-1}x^n = sum_{n=0}^{infty}S_nx^n - sum_{n=0}^{infty}S_{n}x^{n+1} = sum_{n=0}^{infty}S_nx^n(1-x).$$
Now let $M>0.$ Then there is $N$ such $S_n>M$ for $n>N.$ Write the last series as
$$(1-x)left (sum_{n=0}^{N}S_nx^n +sum_{n=N+1}^{infty}S_nx^nright ) > (1-x)left (sum_{n=0}^{N}S_nx^n +Mx^{N+1}frac{1}{1-x}right ).$$
The $liminf_{xto 1^-}$ of the expression on the right equals $0 + M =M.$ We have thus shown the $liminf_{xto 1^-}$ of the left side of $(1)$ is $ge M.$ Since $M$ was arbitrary, this $liminf$ is $infty.$ Thus the limit of left side of $(1)$ is $infty$ as desired.
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
$endgroup$
The following proof that $lim_{xto 1^-}sum_{n=0}^{infty}a_nx^n = infty$ seems simpler to me..
Lemma: Let $A_n=sum_{k=0}^{n}|a_k|.$ Then $sum_{n=0}^{infty}A_nx^n<infty$ for $xin (0,1).$
This follows from the fact that the radius of convergence is $1,$ which is the same as saying $limsup |a_n|^{1/n} =1.$ This is a nice exercise. (A proof of the lemma is now in the comments.)
To prove the main result, let $S_n=sum_{k=0}^{n}a_n;$ set $S_{-1}=0.$ Let $xin (0,1).$ Then
$$tag 1sum_{n=0}^{infty}a_nx^n = sum_{n=0}^{infty}(S_n-S_{n-1})x^n.$$
Now because $|S_n| le A_n,$ the lemma shows we can write the last series as the difference of two convergent series, i.e. as
$$ sum_{n=0}^{infty}S_nx^n - sum_{n=1}^{infty}S_{n-1}x^n = sum_{n=0}^{infty}S_nx^n - sum_{n=0}^{infty}S_{n}x^{n+1} = sum_{n=0}^{infty}S_nx^n(1-x).$$
Now let $M>0.$ Then there is $N$ such $S_n>M$ for $n>N.$ Write the last series as
$$(1-x)left (sum_{n=0}^{N}S_nx^n +sum_{n=N+1}^{infty}S_nx^nright ) > (1-x)left (sum_{n=0}^{N}S_nx^n +Mx^{N+1}frac{1}{1-x}right ).$$
The $liminf_{xto 1^-}$ of the expression on the right equals $0 + M =M.$ We have thus shown the $liminf_{xto 1^-}$ of the left side of $(1)$ is $ge M.$ Since $M$ was arbitrary, this $liminf$ is $infty.$ Thus the limit of left side of $(1)$ is $infty$ as desired.
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
edited Jan 25 at 18:51
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
answered Jan 24 at 21:38
zhw.zhw.
74.2k43175
74.2k43175
$begingroup$
You mean $A_n = sum_{ k = 0}^n mid a_k mid$, no ?
$endgroup$
– Thinking
Jan 24 at 21:58
$begingroup$
@Thinking Yes, thanks. Will correct.
$endgroup$
– zhw.
Jan 24 at 22:00
$begingroup$
You didn't completely correct the typo. I know that if we denote $K_n = sum_{k = 0}^n a_k$ then $sum_{n = 0}^infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$.
$endgroup$
– Thinking
Jan 24 at 22:17
$begingroup$
@Thinking I certainly corrected the typo. It seems you have another question. I'll be back.
$endgroup$
– zhw.
Jan 24 at 22:22
$begingroup$
? So $A_n$ is just $(n+1)mid a_n mid$ ?
$endgroup$
– Thinking
Jan 24 at 22:23
|
show 4 more comments
$begingroup$
You mean $A_n = sum_{ k = 0}^n mid a_k mid$, no ?
$endgroup$
– Thinking
Jan 24 at 21:58
$begingroup$
@Thinking Yes, thanks. Will correct.
$endgroup$
– zhw.
Jan 24 at 22:00
$begingroup$
You didn't completely correct the typo. I know that if we denote $K_n = sum_{k = 0}^n a_k$ then $sum_{n = 0}^infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$.
$endgroup$
– Thinking
Jan 24 at 22:17
$begingroup$
@Thinking I certainly corrected the typo. It seems you have another question. I'll be back.
$endgroup$
– zhw.
Jan 24 at 22:22
$begingroup$
? So $A_n$ is just $(n+1)mid a_n mid$ ?
$endgroup$
– Thinking
Jan 24 at 22:23
$begingroup$
You mean $A_n = sum_{ k = 0}^n mid a_k mid$, no ?
$endgroup$
– Thinking
Jan 24 at 21:58
$begingroup$
You mean $A_n = sum_{ k = 0}^n mid a_k mid$, no ?
$endgroup$
– Thinking
Jan 24 at 21:58
$begingroup$
@Thinking Yes, thanks. Will correct.
$endgroup$
– zhw.
Jan 24 at 22:00
$begingroup$
@Thinking Yes, thanks. Will correct.
$endgroup$
– zhw.
Jan 24 at 22:00
$begingroup$
You didn't completely correct the typo. I know that if we denote $K_n = sum_{k = 0}^n a_k$ then $sum_{n = 0}^infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$.
$endgroup$
– Thinking
Jan 24 at 22:17
$begingroup$
You didn't completely correct the typo. I know that if we denote $K_n = sum_{k = 0}^n a_k$ then $sum_{n = 0}^infty K_n x^n$ converges since it comes from a Cauchy product. But I don't see why $sum_{k = 0}^n A_kx^k$ converges, you don't know if the Cauchy test is going to work on the sequence $a_n$.
$endgroup$
– Thinking
Jan 24 at 22:17
$begingroup$
@Thinking I certainly corrected the typo. It seems you have another question. I'll be back.
$endgroup$
– zhw.
Jan 24 at 22:22
$begingroup$
@Thinking I certainly corrected the typo. It seems you have another question. I'll be back.
$endgroup$
– zhw.
Jan 24 at 22:22
$begingroup$
? So $A_n$ is just $(n+1)mid a_n mid$ ?
$endgroup$
– Thinking
Jan 24 at 22:23
$begingroup$
? So $A_n$ is just $(n+1)mid a_n mid$ ?
$endgroup$
– Thinking
Jan 24 at 22:23
|
show 4 more comments
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$begingroup$
Shouldn't the inequality in your conclusion for positive $a_n$ be the other way round?
$endgroup$
– maxmilgram
Jan 24 at 10:04
$begingroup$
You are searching for an Abel summable sequence.
$endgroup$
– Daniele Tampieri
Jan 24 at 10:41
1
$begingroup$
@maxmilgram No, though it takes a moment to realize that. You have $lim_{xto 1}sum_{n=0}^N a_nx^n=sum_{n=0}^Na_n$ (just polynomials). This means $sum_{n=0}^N a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for some $delta$, corresponding to a chose $epsilon > 0$. With all $a_n$ positive, we get $sum_{n=0}^infty a_nx^ngesum_{n=0}^Na_n - epsilon$ for all $x>1-delta$ for that $delta$. From that follows what's in the post, because $epsilon > 0$ can be chosen freely.
$endgroup$
– Ingix
Jan 24 at 12:25
$begingroup$
Writing $$lim_{x to 1} sum_{n = 0}^infty a_nx^n ne infty$$ is a little strange. Why not ask: Does $$lim_{x to 1^-} sum_{n = 0}^infty a_nx^n =infty?$$
$endgroup$
– zhw.
Jan 24 at 21:52