Mixing
Find the interval of convergence of the series $ sum^{infty}limits_{k=0} ((-1)^k+3)^k(x-1)^k $
$begingroup$
I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}
PROOF
Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}
QUESTION:
Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}
real-analysis sequences-and-series
edited Jan 1 at 12:25
Did
246k23221456
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
$endgroup$
add a comment |
$begingroup$
I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}
PROOF
Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}
QUESTION:
Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}
real-analysis sequences-and-series
edited Jan 1 at 12:25
Did
246k23221456
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
$endgroup$
$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
$endgroup$
– Masacroso
Jan 1 at 11:48
$begingroup$
@Masacroso: Oh oh, that has been my thought too!
$endgroup$
– Omojola Micheal
Jan 1 at 11:54
$begingroup$
@Did: Got your advice!
$endgroup$
– Omojola Micheal
Jan 1 at 12:40
add a comment |
$begingroup$
I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}
PROOF
Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}
QUESTION:
Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}
real-analysis sequences-and-series
edited Jan 1 at 12:25
Did
246k23221456
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
$endgroup$
I wish to find the interval of convergence of the following series
begin{align} sum^{infty}_{k=0} ((-1)^k+3)^k(x-1)^k end{align}
PROOF
Wittingly,
begin{align} left[(-1)^k+3right]^k= begin{cases} 0,&text{if};j=0;\4^j,&text{if};j=2k;\2^j,&text{if};j=2k+1.end{cases} end{align}
Thus,
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}&=|x-1|limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}\&=|x-1|limsup_{ktoinfty} 2^{(2k+1)times frac{1}{k}} \&=4|x-1| end{align}
The series converges absolutely when $limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}<1,$ i.e.,
begin{align} |x-1|< dfrac{1}{4}iff dfrac{3}{4}<x<dfrac{5}{4} end{align}
QUESTION:
Why must begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 2^{2k+1}(x-1)^kright|}end{align}
as stated in the book before me and not
begin{align} limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}= limsup_{ktoinfty} sqrt[k]{left| 4^{2k}(x-1)^kright|};?end{align}
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 1 at 12:25
Did
246k23221456
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
edited Jan 1 at 12:25
Did
246k23221456
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
edited Jan 1 at 12:25
Did
246k23221456
edited Jan 1 at 12:25
Did
246k23221456
edited Jan 1 at 12:25
Did
246k23221456
246k23221456
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
asked Jan 1 at 11:39
Omojola MichealOmojola Micheal
1,753324
1,753324
$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
$endgroup$
– Masacroso
Jan 1 at 11:48
$begingroup$
@Masacroso: Oh oh, that has been my thought too!
$endgroup$
– Omojola Micheal
Jan 1 at 11:54
$begingroup$
@Did: Got your advice!
$endgroup$
– Omojola Micheal
Jan 1 at 12:40
add a comment |
$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
$endgroup$
– Masacroso
Jan 1 at 11:48
$begingroup$
@Masacroso: Oh oh, that has been my thought too!
$endgroup$
– Omojola Micheal
Jan 1 at 11:54
$begingroup$
@Did: Got your advice!
$endgroup$
– Omojola Micheal
Jan 1 at 12:40
$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
$endgroup$
– Masacroso
Jan 1 at 11:48
$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
$endgroup$
– Masacroso
Jan 1 at 11:48
$begingroup$
@Masacroso: Oh oh, that has been my thought too!
$endgroup$
– Omojola Micheal
Jan 1 at 11:54
$begingroup$
@Masacroso: Oh oh, that has been my thought too!
$endgroup$
– Omojola Micheal
Jan 1 at 11:54
$begingroup$
@Did: Got your advice!
$endgroup$
– Omojola Micheal
Jan 1 at 12:40
$begingroup$
@Did: Got your advice!
$endgroup$
– Omojola Micheal
Jan 1 at 12:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Sorry, I edited my work. That equation truly, does not hold!
$endgroup$
– Omojola Micheal
Jan 1 at 12:18
1
$begingroup$
So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
$endgroup$
– Did
Jan 1 at 12:23
$begingroup$
But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
$endgroup$
– Omojola Micheal
Jan 1 at 12:23
$begingroup$
@Did: Sorry, I don't get who you're referring to!
$endgroup$
– Omojola Micheal
Jan 1 at 12:24
$begingroup$
@Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
$endgroup$
– Did
Jan 1 at 12:27
|
show 3 more comments
$begingroup$
Hint
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
$$
and now
$$
sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$
which converges if $|4(x-1)| < 1$ etc. so the result is
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
$$
for $frac 34lt xlt frac 54$
Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Sorry, I edited my work. That equation truly, does not hold!
$endgroup$
– Omojola Micheal
Jan 1 at 12:18
1
$begingroup$
So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
$endgroup$
– Did
Jan 1 at 12:23
$begingroup$
But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
$endgroup$
– Omojola Micheal
Jan 1 at 12:23
$begingroup$
@Did: Sorry, I don't get who you're referring to!
$endgroup$
– Omojola Micheal
Jan 1 at 12:24
$begingroup$
@Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
$endgroup$
– Did
Jan 1 at 12:27
|
show 3 more comments
$begingroup$
Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
Sorry, I edited my work. That equation truly, does not hold!
$endgroup$
– Omojola Micheal
Jan 1 at 12:18
1
$begingroup$
So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
$endgroup$
– Did
Jan 1 at 12:23
$begingroup$
But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
$endgroup$
– Omojola Micheal
Jan 1 at 12:23
$begingroup$
@Did: Sorry, I don't get who you're referring to!
$endgroup$
– Omojola Micheal
Jan 1 at 12:24
$begingroup$
@Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
$endgroup$
– Did
Jan 1 at 12:27
|
show 3 more comments
$begingroup$
Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
Your justification of the equality$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $bigl((-1)^{2k+1}+3bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$sqrt[2k+1]{2^{2k+1}}=2.$$But, since $bigl((-1)^{2k}+3bigr)^{2k}=4^{2k}$ and$$sqrt[2k]{4^{2k}}=4,$$you have$$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=4lvert x-1rvert.$$
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
edited Jan 1 at 12:33
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 12:00
José Carlos SantosJosé Carlos Santos
153k22123225
153k22123225
$begingroup$
Sorry, I edited my work. That equation truly, does not hold!
$endgroup$
– Omojola Micheal
Jan 1 at 12:18
1
$begingroup$
So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
$endgroup$
– Did
Jan 1 at 12:23
$begingroup$
But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
$endgroup$
– Omojola Micheal
Jan 1 at 12:23
$begingroup$
@Did: Sorry, I don't get who you're referring to!
$endgroup$
– Omojola Micheal
Jan 1 at 12:24
$begingroup$
@Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
$endgroup$
– Did
Jan 1 at 12:27
|
show 3 more comments
$begingroup$
Sorry, I edited my work. That equation truly, does not hold!
$endgroup$
– Omojola Micheal
Jan 1 at 12:18
1
$begingroup$
So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
$endgroup$
– Did
Jan 1 at 12:23
$begingroup$
But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
$endgroup$
– Omojola Micheal
Jan 1 at 12:23
$begingroup$
@Did: Sorry, I don't get who you're referring to!
$endgroup$
– Omojola Micheal
Jan 1 at 12:24
$begingroup$
@Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
$endgroup$
– Did
Jan 1 at 12:27
$begingroup$
Sorry, I edited my work. That equation truly, does not hold!
$endgroup$
– Omojola Micheal
Jan 1 at 12:18
$begingroup$
Sorry, I edited my work. That equation truly, does not hold!
$endgroup$
– Omojola Micheal
Jan 1 at 12:18
1
1
$begingroup$
So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
$endgroup$
– Did
Jan 1 at 12:23
$begingroup$
So that, actually, he equality $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lvert x-1rvertlimsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}$$ does hold, as you show, since $$limsup_{ktoinfty}sqrt[k]{ 2^{2k+1}}=4$$
$endgroup$
– Did
Jan 1 at 12:23
$begingroup$
But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
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– Omojola Micheal
Jan 1 at 12:23
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But the power of $(x-1)$ is $k$. How come? From the above, we should have $ limsup_{ktoinfty} sqrt[k]{left|((-1)^k+3)^k(x-1)^kright|}=4^2|x-1|$ ?
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– Omojola Micheal
Jan 1 at 12:23
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@Did: Sorry, I don't get who you're referring to!
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– Omojola Micheal
Jan 1 at 12:24
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@Did: Sorry, I don't get who you're referring to!
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– Omojola Micheal
Jan 1 at 12:24
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@Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
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– Did
Jan 1 at 12:27
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@Mike The OP starts with the assertion that A=B does not hold, then goes on to show that A=C. It happens that, obviously, C=B. Ergo, problem.
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– Did
Jan 1 at 12:27
|
show 3 more comments
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Hint
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
$$
and now
$$
sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$
which converges if $|4(x-1)| < 1$ etc. so the result is
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
$$
for $frac 34lt xlt frac 54$
Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
$endgroup$
add a comment |
$begingroup$
Hint
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
$$
and now
$$
sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$
which converges if $|4(x-1)| < 1$ etc. so the result is
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
$$
for $frac 34lt xlt frac 54$
Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
$endgroup$
add a comment |
$begingroup$
Hint
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
$$
and now
$$
sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$
which converges if $|4(x-1)| < 1$ etc. so the result is
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
$$
for $frac 34lt xlt frac 54$
Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
$endgroup$
Hint
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =lim_{nto infty}left(sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)sum_{k=0}^n 2^{2k}(x-1)^{2k}right)
$$
and now
$$
sum_{k=0}^n 4^{2k}(x-1)^{2k} = frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$
which converges if $|4(x-1)| < 1$ etc. so the result is
$$
sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=frac{2(x-1)}{1-2^2 (x-1)^2}+frac{1}{1-4^2 (x-1)^2}
$$
for $frac 34lt xlt frac 54$
Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
edited Jan 1 at 13:24
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
answered Jan 1 at 12:24
CesareoCesareo
8,4463516
8,4463516
add a comment |
add a comment |
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$begingroup$
if you ask me $$limsup_{ktoinfty}sqrt[k]{((-1)^k+3)^k(x-1)^k}=lim_{ktoinfty}|x-1|sqrt[k]{4^k}=4|x-1|$$ That is, the greater subsequence is when $((-1)^k+3)^k=4^k$. I dont see exactly where comes $2^{2k+1}$
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– Masacroso
Jan 1 at 11:48
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@Masacroso: Oh oh, that has been my thought too!
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– Omojola Micheal
Jan 1 at 11:54
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@Did: Got your advice!
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– Omojola Micheal
Jan 1 at 12:40