Mixing
upper bound on the radius of convergence of a power series
$begingroup$
Let $(a_n)$ be a sequence of complex numbers. Now suppose that $sum na_n$ converge absolutely. Prove that the radius of convergence of $sum_{n = 0}^{infty} a_nx^n$ is $geq 1$.
I don't understand where I am going wrong.
For me it just comes from the fact that
$$forall z, mid z mid leq 1, sum_{n = 0}^infty mid a_n mid mid z^n mid leq sum_{n = 0}^infty mid a_n mid leq sum_{n = 0}^infty n mid a_n mid$$
Hence, for all $z, mid z mid leq 1$ the series $sum a_nz^n$ converges absolutely and hence converges. So for all $z, mid z mid leq 1,$ the series $sum_{n = 0}^infty a_nz^n$ converges, thus the radius of convergence of $x mapsto sum_{n = 0}^infty a_nx^n$ is $1$.
Where is the problem in what I said ?
In my book they are using the fact that $lim_{n to infty} nmid a_n mid to 0$ to prove that the radius of convergence is $1$. So I might be wrong somewhere, since what I said is quite trivial.
Thank you!
real-analysis calculus integration proof-verification power-series
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
Let $(a_n)$ be a sequence of complex numbers. Now suppose that $sum na_n$ converge absolutely. Prove that the radius of convergence of $sum_{n = 0}^{infty} a_nx^n$ is $geq 1$.
I don't understand where I am going wrong.
For me it just comes from the fact that
$$forall z, mid z mid leq 1, sum_{n = 0}^infty mid a_n mid mid z^n mid leq sum_{n = 0}^infty mid a_n mid leq sum_{n = 0}^infty n mid a_n mid$$
Hence, for all $z, mid z mid leq 1$ the series $sum a_nz^n$ converges absolutely and hence converges. So for all $z, mid z mid leq 1,$ the series $sum_{n = 0}^infty a_nz^n$ converges, thus the radius of convergence of $x mapsto sum_{n = 0}^infty a_nx^n$ is $1$.
Where is the problem in what I said ?
In my book they are using the fact that $lim_{n to infty} nmid a_n mid to 0$ to prove that the radius of convergence is $1$. So I might be wrong somewhere, since what I said is quite trivial.
Thank you!
real-analysis calculus integration proof-verification power-series
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
$begingroup$
The title is wrong. It is a lower bound, nor an upper bound.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:46
$begingroup$
Look up and apply the Cauchy-Hadamard Radius Formula, using the fact that $|a_n|^{1/n}=frac {|na_n|^{1/n}}{n^{1/n}}<frac {1}{n^{1/n}}$ for all but finitely many $ nin Bbb Z^+.$
$endgroup$
– DanielWainfleet
Feb 1 at 3:25
add a comment |
$begingroup$
Let $(a_n)$ be a sequence of complex numbers. Now suppose that $sum na_n$ converge absolutely. Prove that the radius of convergence of $sum_{n = 0}^{infty} a_nx^n$ is $geq 1$.
I don't understand where I am going wrong.
For me it just comes from the fact that
$$forall z, mid z mid leq 1, sum_{n = 0}^infty mid a_n mid mid z^n mid leq sum_{n = 0}^infty mid a_n mid leq sum_{n = 0}^infty n mid a_n mid$$
Hence, for all $z, mid z mid leq 1$ the series $sum a_nz^n$ converges absolutely and hence converges. So for all $z, mid z mid leq 1,$ the series $sum_{n = 0}^infty a_nz^n$ converges, thus the radius of convergence of $x mapsto sum_{n = 0}^infty a_nx^n$ is $1$.
Where is the problem in what I said ?
In my book they are using the fact that $lim_{n to infty} nmid a_n mid to 0$ to prove that the radius of convergence is $1$. So I might be wrong somewhere, since what I said is quite trivial.
Thank you!
real-analysis calculus integration proof-verification power-series
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
Let $(a_n)$ be a sequence of complex numbers. Now suppose that $sum na_n$ converge absolutely. Prove that the radius of convergence of $sum_{n = 0}^{infty} a_nx^n$ is $geq 1$.
I don't understand where I am going wrong.
For me it just comes from the fact that
$$forall z, mid z mid leq 1, sum_{n = 0}^infty mid a_n mid mid z^n mid leq sum_{n = 0}^infty mid a_n mid leq sum_{n = 0}^infty n mid a_n mid$$
Hence, for all $z, mid z mid leq 1$ the series $sum a_nz^n$ converges absolutely and hence converges. So for all $z, mid z mid leq 1,$ the series $sum_{n = 0}^infty a_nz^n$ converges, thus the radius of convergence of $x mapsto sum_{n = 0}^infty a_nx^n$ is $1$.
Where is the problem in what I said ?
In my book they are using the fact that $lim_{n to infty} nmid a_n mid to 0$ to prove that the radius of convergence is $1$. So I might be wrong somewhere, since what I said is quite trivial.
Thank you!
real-analysis calculus integration proof-verification power-series
real-analysis calculus integration proof-verification power-series
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
edited Jan 31 at 21:48
J. W. Tanner
4,6191320
4,6191320
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
asked Jan 31 at 21:38
dghkgfzyukzdghkgfzyukz
16612
16612
$begingroup$
The title is wrong. It is a lower bound, nor an upper bound.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:46
$begingroup$
Look up and apply the Cauchy-Hadamard Radius Formula, using the fact that $|a_n|^{1/n}=frac {|na_n|^{1/n}}{n^{1/n}}<frac {1}{n^{1/n}}$ for all but finitely many $ nin Bbb Z^+.$
$endgroup$
– DanielWainfleet
Feb 1 at 3:25
add a comment |
$begingroup$
The title is wrong. It is a lower bound, nor an upper bound.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:46
$begingroup$
Look up and apply the Cauchy-Hadamard Radius Formula, using the fact that $|a_n|^{1/n}=frac {|na_n|^{1/n}}{n^{1/n}}<frac {1}{n^{1/n}}$ for all but finitely many $ nin Bbb Z^+.$
$endgroup$
– DanielWainfleet
Feb 1 at 3:25
$begingroup$
The title is wrong. It is a lower bound, nor an upper bound.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:46
$begingroup$
The title is wrong. It is a lower bound, nor an upper bound.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:46
$begingroup$
Look up and apply the Cauchy-Hadamard Radius Formula, using the fact that $|a_n|^{1/n}=frac {|na_n|^{1/n}}{n^{1/n}}<frac {1}{n^{1/n}}$ for all but finitely many $ nin Bbb Z^+.$
$endgroup$
– DanielWainfleet
Feb 1 at 3:25
$begingroup$
Look up and apply the Cauchy-Hadamard Radius Formula, using the fact that $|a_n|^{1/n}=frac {|na_n|^{1/n}}{n^{1/n}}<frac {1}{n^{1/n}}$ for all but finitely many $ nin Bbb Z^+.$
$endgroup$
– DanielWainfleet
Feb 1 at 3:25
add a comment |
1 Answer
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There is nothing wrong with your approach. Actually, the problem is silly. You would be able to get the some conclusion simply assuming that the series $sum_{n=0}^infty a_n$ converges.
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
This is also what I thought. Thank you, for confirming that what I said is correct !
$endgroup$
– dghkgfzyukz
Jan 31 at 21:42
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
There is nothing wrong with your approach. Actually, the problem is silly. You would be able to get the some conclusion simply assuming that the series $sum_{n=0}^infty a_n$ converges.
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
This is also what I thought. Thank you, for confirming that what I said is correct !
$endgroup$
– dghkgfzyukz
Jan 31 at 21:42
add a comment |
$begingroup$
There is nothing wrong with your approach. Actually, the problem is silly. You would be able to get the some conclusion simply assuming that the series $sum_{n=0}^infty a_n$ converges.
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
This is also what I thought. Thank you, for confirming that what I said is correct !
$endgroup$
– dghkgfzyukz
Jan 31 at 21:42
add a comment |
$begingroup$
There is nothing wrong with your approach. Actually, the problem is silly. You would be able to get the some conclusion simply assuming that the series $sum_{n=0}^infty a_n$ converges.
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
There is nothing wrong with your approach. Actually, the problem is silly. You would be able to get the some conclusion simply assuming that the series $sum_{n=0}^infty a_n$ converges.
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 31 at 21:41
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
$begingroup$
This is also what I thought. Thank you, for confirming that what I said is correct !
$endgroup$
– dghkgfzyukz
Jan 31 at 21:42
add a comment |
$begingroup$
This is also what I thought. Thank you, for confirming that what I said is correct !
$endgroup$
– dghkgfzyukz
Jan 31 at 21:42
$begingroup$
This is also what I thought. Thank you, for confirming that what I said is correct !
$endgroup$
– dghkgfzyukz
Jan 31 at 21:42
$begingroup$
This is also what I thought. Thank you, for confirming that what I said is correct !
$endgroup$
– dghkgfzyukz
Jan 31 at 21:42
add a comment |
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$begingroup$
The title is wrong. It is a lower bound, nor an upper bound.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:46
$begingroup$
Look up and apply the Cauchy-Hadamard Radius Formula, using the fact that $|a_n|^{1/n}=frac {|na_n|^{1/n}}{n^{1/n}}<frac {1}{n^{1/n}}$ for all but finitely many $ nin Bbb Z^+.$
$endgroup$
– DanielWainfleet
Feb 1 at 3:25