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prove or disprove if $lim_{ntoinfty} a_n=0$ then $∑a_n$ converges if and only if $∑a_n+a_{n+1}$ converges...
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Series prove or disprove statement
2 answers
prove or disprove
if $lim_{ntoinfty} a_n=0$ then $∑a_n$ converges if and only if $∑a_n+a_{n+1}$ converges.
I wasn't able to write the indices above and below the ∑. it should be from n=1 to infinity.
I've thought that this statement is incorrect at first, but I found no counterexample.
real-analysis sequences-and-series
edited Jan 17 at 19:57
Bernard
121k740116
asked Jan 17 at 19:49
madam idamadam ida
82
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marked as duplicate by RRL, Martin R, max_zorn, Lord Shark the Unknown, Claude Leibovici sequences-and-series
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Jan 18 at 6:58
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This question already has an answer here:
Series prove or disprove statement
2 answers
prove or disprove
if $lim_{ntoinfty} a_n=0$ then $∑a_n$ converges if and only if $∑a_n+a_{n+1}$ converges.
I wasn't able to write the indices above and below the ∑. it should be from n=1 to infinity.
I've thought that this statement is incorrect at first, but I found no counterexample.
real-analysis sequences-and-series
edited Jan 17 at 19:57
Bernard
121k740116
asked Jan 17 at 19:49
madam idamadam ida
82
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marked as duplicate by RRL, Martin R, max_zorn, Lord Shark the Unknown, Claude Leibovici sequences-and-series
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Jan 18 at 6:58
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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This question already has an answer here:
Series prove or disprove statement
2 answers
prove or disprove
if $lim_{ntoinfty} a_n=0$ then $∑a_n$ converges if and only if $∑a_n+a_{n+1}$ converges.
I wasn't able to write the indices above and below the ∑. it should be from n=1 to infinity.
I've thought that this statement is incorrect at first, but I found no counterexample.
real-analysis sequences-and-series
edited Jan 17 at 19:57
Bernard
121k740116
asked Jan 17 at 19:49
madam idamadam ida
82
$endgroup$
This question already has an answer here:
Series prove or disprove statement
2 answers
prove or disprove
if $lim_{ntoinfty} a_n=0$ then $∑a_n$ converges if and only if $∑a_n+a_{n+1}$ converges.
I wasn't able to write the indices above and below the ∑. it should be from n=1 to infinity.
I've thought that this statement is incorrect at first, but I found no counterexample.
This question already has an answer here:
Series prove or disprove statement
2 answers
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Jan 17 at 19:57
Bernard
121k740116
asked Jan 17 at 19:49
madam idamadam ida
82
edited Jan 17 at 19:57
Bernard
121k740116
asked Jan 17 at 19:49
madam idamadam ida
82
edited Jan 17 at 19:57
Bernard
121k740116
edited Jan 17 at 19:57
Bernard
121k740116
edited Jan 17 at 19:57
Bernard
121k740116
121k740116
asked Jan 17 at 19:49
madam idamadam ida
82
asked Jan 17 at 19:49
madam idamadam ida
82
asked Jan 17 at 19:49
madam idamadam ida
82
82
marked as duplicate by RRL, Martin R, max_zorn, Lord Shark the Unknown, Claude Leibovici sequences-and-series
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Jan 18 at 6:58
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Jan 18 at 6:58
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4 Answers
4
active
oldest
votes
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This is correct. Consider the partial sums for both series:
$$
S_N = sum_{n = 1}^N a_n
$$
and
$$
S_N' = sum_{n = 1}^N (a_n + a_{n+1}) = 2 S_N - a_1 + a_{N+1}
$$
Since $a_n to 0$, the sequence $S_N$ converges if and only if the sequence $S_N'$ converges.
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
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Just consider the partial sums
$$ A_N = sum_{n=1}^{N} a_n,qquad B_N = sum_{n=1}^{N}left(a_n+a_{n+1}right) tag{1}$$
and the relation
$$ B_N = 2 A_N - a_1 + a_{N+1} tag{2}$$
together with the fact that $a_{N+1}to 0$ as $Nto +infty$. In particular ${A_N}_{Ngeq 1}$ is a convergent sequence iff ${B_N}_{Ngeq 1}$ is a convergent sequence.
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
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Let
$S_N = sum_{n=1}^N a_n$
and
$T_N = sum_{n=1}^N (a_n+a_{n+1}).$
Can you find a formula for $S_N$ in terms of $T_N$ and vice versa? (And can you see why doing this would prove the statement?)
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
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Hint Let $S_n$ be the partial sum of $sum a_n$ and $T_n$ be the partial sum of $sum (a_n+a_{n+1})$. Then
$$T_n-2S_n =a_{n+1}-a_1$$
Since $T_n-2S_n$ is convergent to $-a_1$, it follows that $T_n$ is convergent if and only if $S_n$ is convergent [this claim is an immediate consequence of the fact that the sum/difference of two convergent sequences is convergent].
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is correct. Consider the partial sums for both series:
$$
S_N = sum_{n = 1}^N a_n
$$
and
$$
S_N' = sum_{n = 1}^N (a_n + a_{n+1}) = 2 S_N - a_1 + a_{N+1}
$$
Since $a_n to 0$, the sequence $S_N$ converges if and only if the sequence $S_N'$ converges.
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
$endgroup$
add a comment |
$begingroup$
This is correct. Consider the partial sums for both series:
$$
S_N = sum_{n = 1}^N a_n
$$
and
$$
S_N' = sum_{n = 1}^N (a_n + a_{n+1}) = 2 S_N - a_1 + a_{N+1}
$$
Since $a_n to 0$, the sequence $S_N$ converges if and only if the sequence $S_N'$ converges.
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
$endgroup$
add a comment |
$begingroup$
This is correct. Consider the partial sums for both series:
$$
S_N = sum_{n = 1}^N a_n
$$
and
$$
S_N' = sum_{n = 1}^N (a_n + a_{n+1}) = 2 S_N - a_1 + a_{N+1}
$$
Since $a_n to 0$, the sequence $S_N$ converges if and only if the sequence $S_N'$ converges.
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
$endgroup$
This is correct. Consider the partial sums for both series:
$$
S_N = sum_{n = 1}^N a_n
$$
and
$$
S_N' = sum_{n = 1}^N (a_n + a_{n+1}) = 2 S_N - a_1 + a_{N+1}
$$
Since $a_n to 0$, the sequence $S_N$ converges if and only if the sequence $S_N'$ converges.
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
answered Jan 17 at 20:09
Hans EnglerHans Engler
10.5k11836
10.5k11836
add a comment |
add a comment |
$begingroup$
Just consider the partial sums
$$ A_N = sum_{n=1}^{N} a_n,qquad B_N = sum_{n=1}^{N}left(a_n+a_{n+1}right) tag{1}$$
and the relation
$$ B_N = 2 A_N - a_1 + a_{N+1} tag{2}$$
together with the fact that $a_{N+1}to 0$ as $Nto +infty$. In particular ${A_N}_{Ngeq 1}$ is a convergent sequence iff ${B_N}_{Ngeq 1}$ is a convergent sequence.
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
Just consider the partial sums
$$ A_N = sum_{n=1}^{N} a_n,qquad B_N = sum_{n=1}^{N}left(a_n+a_{n+1}right) tag{1}$$
and the relation
$$ B_N = 2 A_N - a_1 + a_{N+1} tag{2}$$
together with the fact that $a_{N+1}to 0$ as $Nto +infty$. In particular ${A_N}_{Ngeq 1}$ is a convergent sequence iff ${B_N}_{Ngeq 1}$ is a convergent sequence.
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
Just consider the partial sums
$$ A_N = sum_{n=1}^{N} a_n,qquad B_N = sum_{n=1}^{N}left(a_n+a_{n+1}right) tag{1}$$
and the relation
$$ B_N = 2 A_N - a_1 + a_{N+1} tag{2}$$
together with the fact that $a_{N+1}to 0$ as $Nto +infty$. In particular ${A_N}_{Ngeq 1}$ is a convergent sequence iff ${B_N}_{Ngeq 1}$ is a convergent sequence.
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
Just consider the partial sums
$$ A_N = sum_{n=1}^{N} a_n,qquad B_N = sum_{n=1}^{N}left(a_n+a_{n+1}right) tag{1}$$
and the relation
$$ B_N = 2 A_N - a_1 + a_{N+1} tag{2}$$
together with the fact that $a_{N+1}to 0$ as $Nto +infty$. In particular ${A_N}_{Ngeq 1}$ is a convergent sequence iff ${B_N}_{Ngeq 1}$ is a convergent sequence.
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 17 at 20:07
Jack D'AurizioJack D'Aurizio
290k33282664
290k33282664
add a comment |
add a comment |
$begingroup$
Let
$S_N = sum_{n=1}^N a_n$
and
$T_N = sum_{n=1}^N (a_n+a_{n+1}).$
Can you find a formula for $S_N$ in terms of $T_N$ and vice versa? (And can you see why doing this would prove the statement?)
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
$endgroup$
add a comment |
$begingroup$
Let
$S_N = sum_{n=1}^N a_n$
and
$T_N = sum_{n=1}^N (a_n+a_{n+1}).$
Can you find a formula for $S_N$ in terms of $T_N$ and vice versa? (And can you see why doing this would prove the statement?)
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
$endgroup$
add a comment |
$begingroup$
Let
$S_N = sum_{n=1}^N a_n$
and
$T_N = sum_{n=1}^N (a_n+a_{n+1}).$
Can you find a formula for $S_N$ in terms of $T_N$ and vice versa? (And can you see why doing this would prove the statement?)
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
$endgroup$
Let
$S_N = sum_{n=1}^N a_n$
and
$T_N = sum_{n=1}^N (a_n+a_{n+1}).$
Can you find a formula for $S_N$ in terms of $T_N$ and vice versa? (And can you see why doing this would prove the statement?)
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
answered Jan 17 at 20:08
Jordan GreenJordan Green
1,119310
1,119310
add a comment |
add a comment |
$begingroup$
Hint Let $S_n$ be the partial sum of $sum a_n$ and $T_n$ be the partial sum of $sum (a_n+a_{n+1})$. Then
$$T_n-2S_n =a_{n+1}-a_1$$
Since $T_n-2S_n$ is convergent to $-a_1$, it follows that $T_n$ is convergent if and only if $S_n$ is convergent [this claim is an immediate consequence of the fact that the sum/difference of two convergent sequences is convergent].
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Hint Let $S_n$ be the partial sum of $sum a_n$ and $T_n$ be the partial sum of $sum (a_n+a_{n+1})$. Then
$$T_n-2S_n =a_{n+1}-a_1$$
Since $T_n-2S_n$ is convergent to $-a_1$, it follows that $T_n$ is convergent if and only if $S_n$ is convergent [this claim is an immediate consequence of the fact that the sum/difference of two convergent sequences is convergent].
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Hint Let $S_n$ be the partial sum of $sum a_n$ and $T_n$ be the partial sum of $sum (a_n+a_{n+1})$. Then
$$T_n-2S_n =a_{n+1}-a_1$$
Since $T_n-2S_n$ is convergent to $-a_1$, it follows that $T_n$ is convergent if and only if $S_n$ is convergent [this claim is an immediate consequence of the fact that the sum/difference of two convergent sequences is convergent].
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
$endgroup$
Hint Let $S_n$ be the partial sum of $sum a_n$ and $T_n$ be the partial sum of $sum (a_n+a_{n+1})$. Then
$$T_n-2S_n =a_{n+1}-a_1$$
Since $T_n-2S_n$ is convergent to $-a_1$, it follows that $T_n$ is convergent if and only if $S_n$ is convergent [this claim is an immediate consequence of the fact that the sum/difference of two convergent sequences is convergent].
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
answered Jan 17 at 20:09
N. S.N. S.
104k7114209
104k7114209
add a comment |
add a comment |
