Determine if this specific sequence is a Cauchy sequence

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I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$



So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$



What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $



$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:



$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $



From here on, its not so clear to me as to how to proceed.
What should be my next steps?






calculus sequences-and-series limits cauchy-sequences







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  • 1




    Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
    – Yanko
    Dec 9 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 at 10:51














up vote
3
down vote

favorite












I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$



So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$



What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $



$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:



$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $



From here on, its not so clear to me as to how to proceed.
What should be my next steps?






calculus sequences-and-series limits cauchy-sequences







share|cite|improve this question



















  • 1




    Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
    – Yanko
    Dec 9 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 at 10:51












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$



So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$



What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $



$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:



$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $



From here on, its not so clear to me as to how to proceed.
What should be my next steps?






calculus sequences-and-series limits cauchy-sequences







share|cite|improve this question















I have the following sequence:
$$a_n =sum_k = 1^n (-1)^b_k 1over k^2 $$
And the hint is that I have to prove that:
$$ 1over k^2 < 1over k-1 - 1over k $$



So assuming $m>n$, I have to prove that:
$$forall epsilon >0, exists N in mathbbN,$$ so that $$ forall m,n > N Rightarrow lvert a_m - a_nrvert < epsilon $$



What I gathered so far:
$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert $



$b_k$ is a sequence of natural numbers $1,2,3.....$, so in absolute value, $(-1)^b_k $ is $1$.
Therefore:



$ lvert a_m - a_nrvert = lvert sum_k = n+1^m (-1)^b_k 1over k^2rvert leq sum_k = n+1^m 1over k^2. $



From here on, its not so clear to me as to how to proceed.
What should be my next steps?






calculus sequences-and-series limits cauchy-sequences




calculus sequences-and-series limits cauchy-sequences






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 10:51









Jonas Lenz

505212




505212










asked Dec 9 at 10:28









Tegernako

616




616







  • 1




    Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
    – Yanko
    Dec 9 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 at 10:51












  • 1




    Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
    – Yanko
    Dec 9 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 at 10:51







1




1




Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50




Note that (as suggested in one of the edits) the last equality in the last line should be an inequality $leq$.
– Yanko
Dec 9 at 10:50












Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51




Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 at 10:51










2 Answers
2






active

oldest

votes

















up vote
5
down vote



accepted










You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that



$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer






















  • Exactly! Thanks.
    – Tegernako
    Dec 9 at 10:41

















up vote
5
down vote













Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$






share|cite|improve this answer


















  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 at 10:44







  • 1




    In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 at 10:46







  • 1




    @JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
    – Yanko
    Dec 9 at 10:49










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that



$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer






















  • Exactly! Thanks.
    – Tegernako
    Dec 9 at 10:41














up vote
5
down vote



accepted










You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that



$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer






















  • Exactly! Thanks.
    – Tegernako
    Dec 9 at 10:41












up vote
5
down vote



accepted







up vote
5
down vote



accepted






You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that



$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer














You're almost done. Since $frac1k^2 leq frac1k-1-frac1k$ you have that



$$sum_k = n+1^m 1over k^2leqsum_k=n+1^m left[frac1k-1-frac1kright]$$ This is a telescoping series which is equal to $frac1n-frac1m$. It converges to zero as $n,mrightarrow 0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 at 10:51

























answered Dec 9 at 10:35









Yanko

5,760723




5,760723











  • Exactly! Thanks.
    – Tegernako
    Dec 9 at 10:41
















  • Exactly! Thanks.
    – Tegernako
    Dec 9 at 10:41















Exactly! Thanks.
– Tegernako
Dec 9 at 10:41




Exactly! Thanks.
– Tegernako
Dec 9 at 10:41










up vote
5
down vote













Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$






share|cite|improve this answer


















  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 at 10:44







  • 1




    In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 at 10:46







  • 1




    @JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
    – Yanko
    Dec 9 at 10:49














up vote
5
down vote













Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$






share|cite|improve this answer


















  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 at 10:44







  • 1




    In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 at 10:46







  • 1




    @JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
    – Yanko
    Dec 9 at 10:49












up vote
5
down vote










up vote
5
down vote









Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$






share|cite|improve this answer














Now, use the fact that$$sum_k=n+1^mfrac1k^2<sum_k=n+1^mfrac1k-frac1k-1=frac1n-frac1m.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 at 10:39

























answered Dec 9 at 10:36









José Carlos Santos

147k22117218




147k22117218







  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 at 10:44







  • 1




    In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 at 10:46







  • 1




    @JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
    – Yanko
    Dec 9 at 10:49












  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 at 10:44







  • 1




    In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 at 10:46







  • 1




    @JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
    – Yanko
    Dec 9 at 10:49







1




1




I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39




I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 at 10:39












Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40




Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 at 10:40




1




1




Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44





Why do we need the hint? Can't we simply say that as $sum_k=1^infty frac1k^2$ is convergent, we know that that $sum_k=n^infty frac1k^2$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 at 10:44





1




1




In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46





In order to prove that, for each $varepsilon>0$, $sum_k=n+1^mfrac1k^2<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 at 10:46





1




1




@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49




@JonasLenz I guess that is how one shows that $sum_k=1^infty frac1k^2$ convergent.
– Yanko
Dec 9 at 10:49

















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