rearrangement of the alternating harmonic series [duplicate]
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This question already has an answer here:
Why does the order of summation of the terms of an infinite series influence its value?
3 answers
The alternating harmonic series $a:=sum_{n=1}^{infty}{frac{(-1)^{n-1}}n}$ converges by the Leibnitz-criteria. Now consider the following rearrangement of this series given by:
begin{align}
S:=1+frac{1}{3}-frac{1}{2}-frac{1}{4}+frac{1}{5}+frac{1}{7}-...+frac{1}{4n-3}+frac{1}{4n-1}-frac{1}{4n-2}-frac{1}{4n}+...
end{align}
I want to see that this series converges aswell and has the same value $a$. I am stuck with this one. Does someone have a hint how to start this?
Edit: this not a duplicate to the question in the link below since I am asking on how to get the limit value of this series.
real-analysis sequences-and-series convergence
asked Dec 15 '18 at 11:42
JulesJules
1087
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marked as duplicate by TheSimpliFire, Did, BigbearZzz, kjetil b halvorsen, the_candyman Dec 16 '18 at 20:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why does the order of summation of the terms of an infinite series influence its value?
3 answers
The alternating harmonic series $a:=sum_{n=1}^{infty}{frac{(-1)^{n-1}}n}$ converges by the Leibnitz-criteria. Now consider the following rearrangement of this series given by:
begin{align}
S:=1+frac{1}{3}-frac{1}{2}-frac{1}{4}+frac{1}{5}+frac{1}{7}-...+frac{1}{4n-3}+frac{1}{4n-1}-frac{1}{4n-2}-frac{1}{4n}+...
end{align}
I want to see that this series converges aswell and has the same value $a$. I am stuck with this one. Does someone have a hint how to start this?
Edit: this not a duplicate to the question in the link below since I am asking on how to get the limit value of this series.
real-analysis sequences-and-series convergence
asked Dec 15 '18 at 11:42
JulesJules
1087
$endgroup$
marked as duplicate by TheSimpliFire, Did, BigbearZzz, kjetil b halvorsen, the_candyman Dec 16 '18 at 20:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
@TheSimpliFire I don't think it is an duplicate because the question in your link is way more general
$endgroup$
– Jules
Dec 15 '18 at 11:46
$begingroup$
@TheSimpliFire my series is a diffrent one tho....
$endgroup$
– Jules
Dec 15 '18 at 11:52
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It uses the same principle. Try it with your series.
$endgroup$
– TheSimpliFire
Dec 15 '18 at 13:46
1
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This is a harmless rearrangement as the terms move by a constant displacement. Trouble occurs when terms get "rejected to infinity".
$endgroup$
– Yves Daoust
Dec 15 '18 at 20:10
add a comment |
$begingroup$
This question already has an answer here:
Why does the order of summation of the terms of an infinite series influence its value?
3 answers
The alternating harmonic series $a:=sum_{n=1}^{infty}{frac{(-1)^{n-1}}n}$ converges by the Leibnitz-criteria. Now consider the following rearrangement of this series given by:
begin{align}
S:=1+frac{1}{3}-frac{1}{2}-frac{1}{4}+frac{1}{5}+frac{1}{7}-...+frac{1}{4n-3}+frac{1}{4n-1}-frac{1}{4n-2}-frac{1}{4n}+...
end{align}
I want to see that this series converges aswell and has the same value $a$. I am stuck with this one. Does someone have a hint how to start this?
Edit: this not a duplicate to the question in the link below since I am asking on how to get the limit value of this series.
real-analysis sequences-and-series convergence
asked Dec 15 '18 at 11:42
JulesJules
1087
$endgroup$
This question already has an answer here:
Why does the order of summation of the terms of an infinite series influence its value?
3 answers
The alternating harmonic series $a:=sum_{n=1}^{infty}{frac{(-1)^{n-1}}n}$ converges by the Leibnitz-criteria. Now consider the following rearrangement of this series given by:
begin{align}
S:=1+frac{1}{3}-frac{1}{2}-frac{1}{4}+frac{1}{5}+frac{1}{7}-...+frac{1}{4n-3}+frac{1}{4n-1}-frac{1}{4n-2}-frac{1}{4n}+...
end{align}
I want to see that this series converges aswell and has the same value $a$. I am stuck with this one. Does someone have a hint how to start this?
Edit: this not a duplicate to the question in the link below since I am asking on how to get the limit value of this series.
This question already has an answer here:
Why does the order of summation of the terms of an infinite series influence its value?
3 answers
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked Dec 15 '18 at 11:42
JulesJules
1087
asked Dec 15 '18 at 11:42
JulesJules
1087
edited Dec 15 '18 at 11:48
Jules
asked Dec 15 '18 at 11:42
JulesJules
1087
asked Dec 15 '18 at 11:42
JulesJules
1087
asked Dec 15 '18 at 11:42
JulesJules
1087
1087
marked as duplicate by TheSimpliFire, Did, BigbearZzz, kjetil b halvorsen, the_candyman Dec 16 '18 at 20:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by TheSimpliFire, Did, BigbearZzz, kjetil b halvorsen, the_candyman Dec 16 '18 at 20:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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@TheSimpliFire I don't think it is an duplicate because the question in your link is way more general
$endgroup$
– Jules
Dec 15 '18 at 11:46
$begingroup$
@TheSimpliFire my series is a diffrent one tho....
$endgroup$
– Jules
Dec 15 '18 at 11:52
$begingroup$
It uses the same principle. Try it with your series.
$endgroup$
– TheSimpliFire
Dec 15 '18 at 13:46
1
$begingroup$
This is a harmless rearrangement as the terms move by a constant displacement. Trouble occurs when terms get "rejected to infinity".
$endgroup$
– Yves Daoust
Dec 15 '18 at 20:10
add a comment |
$begingroup$
@TheSimpliFire I don't think it is an duplicate because the question in your link is way more general
$endgroup$
– Jules
Dec 15 '18 at 11:46
$begingroup$
@TheSimpliFire my series is a diffrent one tho....
$endgroup$
– Jules
Dec 15 '18 at 11:52
$begingroup$
It uses the same principle. Try it with your series.
$endgroup$
– TheSimpliFire
Dec 15 '18 at 13:46
1
$begingroup$
This is a harmless rearrangement as the terms move by a constant displacement. Trouble occurs when terms get "rejected to infinity".
$endgroup$
– Yves Daoust
Dec 15 '18 at 20:10
$begingroup$
@TheSimpliFire I don't think it is an duplicate because the question in your link is way more general
$endgroup$
– Jules
Dec 15 '18 at 11:46
$begingroup$
@TheSimpliFire I don't think it is an duplicate because the question in your link is way more general
$endgroup$
– Jules
Dec 15 '18 at 11:46
$begingroup$
@TheSimpliFire my series is a diffrent one tho....
$endgroup$
– Jules
Dec 15 '18 at 11:52
$begingroup$
@TheSimpliFire my series is a diffrent one tho....
$endgroup$
– Jules
Dec 15 '18 at 11:52
$begingroup$
It uses the same principle. Try it with your series.
$endgroup$
– TheSimpliFire
Dec 15 '18 at 13:46
$begingroup$
It uses the same principle. Try it with your series.
$endgroup$
– TheSimpliFire
Dec 15 '18 at 13:46
1
1
$begingroup$
This is a harmless rearrangement as the terms move by a constant displacement. Trouble occurs when terms get "rejected to infinity".
$endgroup$
– Yves Daoust
Dec 15 '18 at 20:10
$begingroup$
This is a harmless rearrangement as the terms move by a constant displacement. Trouble occurs when terms get "rejected to infinity".
$endgroup$
– Yves Daoust
Dec 15 '18 at 20:10
add a comment |
1 Answer
1
active
oldest
votes
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Hint: Let $S_n$ be the $n$th partial sum of the alternating harmonic series, and let $T_n$ be the $n$th partial sum of the given rearrangement. Then $S_{4n}= T_{4n}$ for all $n.$ How far from $T_{4n}$ can $T_{4n+1},T_{4n+2},T_{4n+3}$ be?
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
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$begingroup$
Not very far since the sequence of summands of the rearranged series form a sequence that converges to zero! Thanks, I got it!
$endgroup$
– Jules
Dec 16 '18 at 23:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $S_n$ be the $n$th partial sum of the alternating harmonic series, and let $T_n$ be the $n$th partial sum of the given rearrangement. Then $S_{4n}= T_{4n}$ for all $n.$ How far from $T_{4n}$ can $T_{4n+1},T_{4n+2},T_{4n+3}$ be?
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
$endgroup$
$begingroup$
Not very far since the sequence of summands of the rearranged series form a sequence that converges to zero! Thanks, I got it!
$endgroup$
– Jules
Dec 16 '18 at 23:12
add a comment |
$begingroup$
Hint: Let $S_n$ be the $n$th partial sum of the alternating harmonic series, and let $T_n$ be the $n$th partial sum of the given rearrangement. Then $S_{4n}= T_{4n}$ for all $n.$ How far from $T_{4n}$ can $T_{4n+1},T_{4n+2},T_{4n+3}$ be?
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
$endgroup$
$begingroup$
Not very far since the sequence of summands of the rearranged series form a sequence that converges to zero! Thanks, I got it!
$endgroup$
– Jules
Dec 16 '18 at 23:12
add a comment |
$begingroup$
Hint: Let $S_n$ be the $n$th partial sum of the alternating harmonic series, and let $T_n$ be the $n$th partial sum of the given rearrangement. Then $S_{4n}= T_{4n}$ for all $n.$ How far from $T_{4n}$ can $T_{4n+1},T_{4n+2},T_{4n+3}$ be?
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
$endgroup$
Hint: Let $S_n$ be the $n$th partial sum of the alternating harmonic series, and let $T_n$ be the $n$th partial sum of the given rearrangement. Then $S_{4n}= T_{4n}$ for all $n.$ How far from $T_{4n}$ can $T_{4n+1},T_{4n+2},T_{4n+3}$ be?
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
answered Dec 15 '18 at 19:19
zhw.zhw.
72.9k43175
72.9k43175
$begingroup$
Not very far since the sequence of summands of the rearranged series form a sequence that converges to zero! Thanks, I got it!
$endgroup$
– Jules
Dec 16 '18 at 23:12
add a comment |
$begingroup$
Not very far since the sequence of summands of the rearranged series form a sequence that converges to zero! Thanks, I got it!
$endgroup$
– Jules
Dec 16 '18 at 23:12
$begingroup$
Not very far since the sequence of summands of the rearranged series form a sequence that converges to zero! Thanks, I got it!
$endgroup$
– Jules
Dec 16 '18 at 23:12
$begingroup$
Not very far since the sequence of summands of the rearranged series form a sequence that converges to zero! Thanks, I got it!
$endgroup$
– Jules
Dec 16 '18 at 23:12
add a comment |
$begingroup$
@TheSimpliFire I don't think it is an duplicate because the question in your link is way more general
$endgroup$
– Jules
Dec 15 '18 at 11:46
$begingroup$
@TheSimpliFire my series is a diffrent one tho....
$endgroup$
– Jules
Dec 15 '18 at 11:52
$begingroup$
It uses the same principle. Try it with your series.
$endgroup$
– TheSimpliFire
Dec 15 '18 at 13:46
1
$begingroup$
This is a harmless rearrangement as the terms move by a constant displacement. Trouble occurs when terms get "rejected to infinity".
$endgroup$
– Yves Daoust
Dec 15 '18 at 20:10