Mixing
On the convergence of a two-sided series
$begingroup$
Background
Let $left{ {{a_n}} right}_{ - infty }^infty $ be a two sided sequence (is there a more proper term?) of complex numbers.
As far as I know (please correct me if I am wrong) we say that $$sumlimits_{n = - infty }^infty {{a_n}} $$ converges if and only if the following two series converge:
$$sumlimits_{n = - infty }^{ - 1} {{a_n}} ,sumlimits_{n = 0}^infty {{a_n}} $$
We know that the existence of $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {{a_n}} $ does not imply the convergence of $sumlimits_{n = - infty }^infty {{a_n}} $
since for example $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N n $ exists and the limit is zero (it is in fact the zero sequence, which limit is also zero) but the series $sumlimits_{n = 0}^infty n $ does not converge, therefore $sumlimits_{n = - infty }^infty {{a_n}} $ diverges.
My question is the following:
(1) Let's say $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists. Does that imply that $sumlimits_{n = - infty }^infty {{a_n}} $ converges? converges absolutely?
I think so and I'll try to prove it.
My attempt:
Let $varepsilon > 0$.
Since the limit $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists, by the Cauchy criterion we have a natural number ${N_0} $ such that for all $N > {N_0}$ and for all natural $p$ the following takes hold:
$sumlimits_{n = - N + p}^{N + p} {left| {{a_n}} right|} - sumlimits_{n = - N}^N {left| {{a_n}} right|} = sumlimits_{ - left( {N + p} right)}^{ - left( {N + 1} right)} {left| {{a_n}} right|} + sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
But that implies
(1) $sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = 0}^infty {{a_n}} $ converges absolutely.
(2) $sumlimits_{-(N + p)}^{-(N + 1)} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = -infty}^{-1} {{a_n}} $ converges absolutely.
Thus $sumlimits_{n = - infty }^infty {{a_n}} $ converge absolutely and therefore converges.
Am I correct?
sequences-and-series proof-verification absolute-convergence
asked Jan 3 at 5:56
zokomokozokomoko
162214
$endgroup$
add a comment |
$begingroup$
Background
Let $left{ {{a_n}} right}_{ - infty }^infty $ be a two sided sequence (is there a more proper term?) of complex numbers.
As far as I know (please correct me if I am wrong) we say that $$sumlimits_{n = - infty }^infty {{a_n}} $$ converges if and only if the following two series converge:
$$sumlimits_{n = - infty }^{ - 1} {{a_n}} ,sumlimits_{n = 0}^infty {{a_n}} $$
We know that the existence of $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {{a_n}} $ does not imply the convergence of $sumlimits_{n = - infty }^infty {{a_n}} $
since for example $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N n $ exists and the limit is zero (it is in fact the zero sequence, which limit is also zero) but the series $sumlimits_{n = 0}^infty n $ does not converge, therefore $sumlimits_{n = - infty }^infty {{a_n}} $ diverges.
My question is the following:
(1) Let's say $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists. Does that imply that $sumlimits_{n = - infty }^infty {{a_n}} $ converges? converges absolutely?
I think so and I'll try to prove it.
My attempt:
Let $varepsilon > 0$.
Since the limit $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists, by the Cauchy criterion we have a natural number ${N_0} $ such that for all $N > {N_0}$ and for all natural $p$ the following takes hold:
$sumlimits_{n = - N + p}^{N + p} {left| {{a_n}} right|} - sumlimits_{n = - N}^N {left| {{a_n}} right|} = sumlimits_{ - left( {N + p} right)}^{ - left( {N + 1} right)} {left| {{a_n}} right|} + sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
But that implies
(1) $sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = 0}^infty {{a_n}} $ converges absolutely.
(2) $sumlimits_{-(N + p)}^{-(N + 1)} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = -infty}^{-1} {{a_n}} $ converges absolutely.
Thus $sumlimits_{n = - infty }^infty {{a_n}} $ converge absolutely and therefore converges.
Am I correct?
sequences-and-series proof-verification absolute-convergence
asked Jan 3 at 5:56
zokomokozokomoko
162214
$endgroup$
3
$begingroup$
Your proof is fine but I suggest using an upper bound instead of Cauchy property. $lim sum_{-N}^{N}|a_n|$ exists iff there is a finite constant $M$ such that $sum_{-N}^{N}|a_n| leq M$ for all $n$ and the rest should be clear from this.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:15
1
$begingroup$
Thanks, great suggestion!
$endgroup$
– zokomoko
Jan 3 at 6:20
add a comment |
$begingroup$
Background
Let $left{ {{a_n}} right}_{ - infty }^infty $ be a two sided sequence (is there a more proper term?) of complex numbers.
As far as I know (please correct me if I am wrong) we say that $$sumlimits_{n = - infty }^infty {{a_n}} $$ converges if and only if the following two series converge:
$$sumlimits_{n = - infty }^{ - 1} {{a_n}} ,sumlimits_{n = 0}^infty {{a_n}} $$
We know that the existence of $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {{a_n}} $ does not imply the convergence of $sumlimits_{n = - infty }^infty {{a_n}} $
since for example $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N n $ exists and the limit is zero (it is in fact the zero sequence, which limit is also zero) but the series $sumlimits_{n = 0}^infty n $ does not converge, therefore $sumlimits_{n = - infty }^infty {{a_n}} $ diverges.
My question is the following:
(1) Let's say $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists. Does that imply that $sumlimits_{n = - infty }^infty {{a_n}} $ converges? converges absolutely?
I think so and I'll try to prove it.
My attempt:
Let $varepsilon > 0$.
Since the limit $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists, by the Cauchy criterion we have a natural number ${N_0} $ such that for all $N > {N_0}$ and for all natural $p$ the following takes hold:
$sumlimits_{n = - N + p}^{N + p} {left| {{a_n}} right|} - sumlimits_{n = - N}^N {left| {{a_n}} right|} = sumlimits_{ - left( {N + p} right)}^{ - left( {N + 1} right)} {left| {{a_n}} right|} + sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
But that implies
(1) $sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = 0}^infty {{a_n}} $ converges absolutely.
(2) $sumlimits_{-(N + p)}^{-(N + 1)} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = -infty}^{-1} {{a_n}} $ converges absolutely.
Thus $sumlimits_{n = - infty }^infty {{a_n}} $ converge absolutely and therefore converges.
Am I correct?
sequences-and-series proof-verification absolute-convergence
asked Jan 3 at 5:56
zokomokozokomoko
162214
$endgroup$
Background
Let $left{ {{a_n}} right}_{ - infty }^infty $ be a two sided sequence (is there a more proper term?) of complex numbers.
As far as I know (please correct me if I am wrong) we say that $$sumlimits_{n = - infty }^infty {{a_n}} $$ converges if and only if the following two series converge:
$$sumlimits_{n = - infty }^{ - 1} {{a_n}} ,sumlimits_{n = 0}^infty {{a_n}} $$
We know that the existence of $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {{a_n}} $ does not imply the convergence of $sumlimits_{n = - infty }^infty {{a_n}} $
since for example $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N n $ exists and the limit is zero (it is in fact the zero sequence, which limit is also zero) but the series $sumlimits_{n = 0}^infty n $ does not converge, therefore $sumlimits_{n = - infty }^infty {{a_n}} $ diverges.
My question is the following:
(1) Let's say $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists. Does that imply that $sumlimits_{n = - infty }^infty {{a_n}} $ converges? converges absolutely?
I think so and I'll try to prove it.
My attempt:
Let $varepsilon > 0$.
Since the limit $mathop {lim }limits_{N to infty } sumlimits_{n = - N}^N {left| {{a_n}} right|} $ exists, by the Cauchy criterion we have a natural number ${N_0} $ such that for all $N > {N_0}$ and for all natural $p$ the following takes hold:
$sumlimits_{n = - N + p}^{N + p} {left| {{a_n}} right|} - sumlimits_{n = - N}^N {left| {{a_n}} right|} = sumlimits_{ - left( {N + p} right)}^{ - left( {N + 1} right)} {left| {{a_n}} right|} + sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
But that implies
(1) $sumlimits_{N + 1}^{N + p} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = 0}^infty {{a_n}} $ converges absolutely.
(2) $sumlimits_{-(N + p)}^{-(N + 1)} {left| {{a_n}} right|} < varepsilon $
so by Cauchy's criterion the series $sumlimits_{n = -infty}^{-1} {{a_n}} $ converges absolutely.
Thus $sumlimits_{n = - infty }^infty {{a_n}} $ converge absolutely and therefore converges.
Am I correct?
sequences-and-series proof-verification absolute-convergence
sequences-and-series proof-verification absolute-convergence
asked Jan 3 at 5:56
zokomokozokomoko
162214
asked Jan 3 at 5:56
zokomokozokomoko
162214
asked Jan 3 at 5:56
zokomokozokomoko
162214
asked Jan 3 at 5:56
zokomokozokomoko
162214
asked Jan 3 at 5:56
zokomokozokomoko
162214
162214
3
$begingroup$
Your proof is fine but I suggest using an upper bound instead of Cauchy property. $lim sum_{-N}^{N}|a_n|$ exists iff there is a finite constant $M$ such that $sum_{-N}^{N}|a_n| leq M$ for all $n$ and the rest should be clear from this.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:15
1
$begingroup$
Thanks, great suggestion!
$endgroup$
– zokomoko
Jan 3 at 6:20
add a comment |
3
$begingroup$
Your proof is fine but I suggest using an upper bound instead of Cauchy property. $lim sum_{-N}^{N}|a_n|$ exists iff there is a finite constant $M$ such that $sum_{-N}^{N}|a_n| leq M$ for all $n$ and the rest should be clear from this.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:15
1
$begingroup$
Thanks, great suggestion!
$endgroup$
– zokomoko
Jan 3 at 6:20
3
3
$begingroup$
Your proof is fine but I suggest using an upper bound instead of Cauchy property. $lim sum_{-N}^{N}|a_n|$ exists iff there is a finite constant $M$ such that $sum_{-N}^{N}|a_n| leq M$ for all $n$ and the rest should be clear from this.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:15
$begingroup$
Your proof is fine but I suggest using an upper bound instead of Cauchy property. $lim sum_{-N}^{N}|a_n|$ exists iff there is a finite constant $M$ such that $sum_{-N}^{N}|a_n| leq M$ for all $n$ and the rest should be clear from this.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:15
1
1
$begingroup$
Thanks, great suggestion!
$endgroup$
– zokomoko
Jan 3 at 6:20
$begingroup$
Thanks, great suggestion!
$endgroup$
– zokomoko
Jan 3 at 6:20
add a comment |
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3
$begingroup$
Your proof is fine but I suggest using an upper bound instead of Cauchy property. $lim sum_{-N}^{N}|a_n|$ exists iff there is a finite constant $M$ such that $sum_{-N}^{N}|a_n| leq M$ for all $n$ and the rest should be clear from this.
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:15
1
$begingroup$
Thanks, great suggestion!
$endgroup$
– zokomoko
Jan 3 at 6:20