Determine if this specific sequence is a Cauchy sequence












3














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:
$$forall epsilon >0, exists N in mathbb{N},$$ 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^2}rvert $



$b_k$ 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^2}rvert 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 '18 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 '18 at 10:51
















3














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:
$$forall epsilon >0, exists N in mathbb{N},$$ 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^2}rvert $



$b_k$ 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^2}rvert 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 '18 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 '18 at 10:51














3












3








3







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:
$$forall epsilon >0, exists N in mathbb{N},$$ 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^2}rvert $



$b_k$ 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^2}rvert 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:
$$forall epsilon >0, exists N in mathbb{N},$$ 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^2}rvert $



$b_k$ 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^2}rvert 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 '18 at 10:51









Jonas Lenz

530212




530212










asked Dec 9 '18 at 10:28









Tegernako

756




756








  • 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 '18 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 '18 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 '18 at 10:50










  • Just saw it, fixed, thanks a lot :)
    – Tegernako
    Dec 9 '18 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 '18 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 '18 at 10:50












Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 '18 at 10:51




Just saw it, fixed, thanks a lot :)
– Tegernako
Dec 9 '18 at 10:51










2 Answers
2






active

oldest

votes


















5














You're almost done. Since $frac{1}{k^2} leq frac{1}{k-1}-frac{1}{k}$ you have that



$$sum_{k = n+1}^m {1over k^2}leqsum_{k=n+1}^m left[frac{1}{k-1}-frac{1}{k}right]$$ This is a telescoping series which is equal to $frac{1}{n}-frac{1}{m}$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer























  • Exactly! Thanks.
    – Tegernako
    Dec 9 '18 at 10:41



















5














Now, use the fact that$$sum_{k=n+1}^mfrac1{k^2}<sum_{k=n+1}^mfrac1k-frac1{k-1}=frac1n-frac1m.$$






share|cite|improve this answer



















  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 '18 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 '18 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_{k=1}^infty frac{1}{k^2}$ is convergent, we know that that $sum_{k=n}^infty frac{1}{k^2}$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 '18 at 10:44








  • 1




    In order to prove that, for each $varepsilon>0$, $sum_{k=n+1}^mfrac1{k^2}<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 '18 at 10:46








  • 1




    @JonasLenz I guess that is how one shows that $sum_{k=1}^infty frac{1}{k^2}$ convergent.
    – Yanko
    Dec 9 '18 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









5














You're almost done. Since $frac{1}{k^2} leq frac{1}{k-1}-frac{1}{k}$ you have that



$$sum_{k = n+1}^m {1over k^2}leqsum_{k=n+1}^m left[frac{1}{k-1}-frac{1}{k}right]$$ This is a telescoping series which is equal to $frac{1}{n}-frac{1}{m}$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer























  • Exactly! Thanks.
    – Tegernako
    Dec 9 '18 at 10:41
















5














You're almost done. Since $frac{1}{k^2} leq frac{1}{k-1}-frac{1}{k}$ you have that



$$sum_{k = n+1}^m {1over k^2}leqsum_{k=n+1}^m left[frac{1}{k-1}-frac{1}{k}right]$$ This is a telescoping series which is equal to $frac{1}{n}-frac{1}{m}$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer























  • Exactly! Thanks.
    – Tegernako
    Dec 9 '18 at 10:41














5












5








5






You're almost done. Since $frac{1}{k^2} leq frac{1}{k-1}-frac{1}{k}$ you have that



$$sum_{k = n+1}^m {1over k^2}leqsum_{k=n+1}^m left[frac{1}{k-1}-frac{1}{k}right]$$ This is a telescoping series which is equal to $frac{1}{n}-frac{1}{m}$. It converges to zero as $n,mrightarrow 0$.






share|cite|improve this answer














You're almost done. Since $frac{1}{k^2} leq frac{1}{k-1}-frac{1}{k}$ you have that



$$sum_{k = n+1}^m {1over k^2}leqsum_{k=n+1}^m left[frac{1}{k-1}-frac{1}{k}right]$$ This is a telescoping series which is equal to $frac{1}{n}-frac{1}{m}$. 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 '18 at 10:51

























answered Dec 9 '18 at 10:35









Yanko

6,124724




6,124724












  • Exactly! Thanks.
    – Tegernako
    Dec 9 '18 at 10:41


















  • Exactly! Thanks.
    – Tegernako
    Dec 9 '18 at 10:41
















Exactly! Thanks.
– Tegernako
Dec 9 '18 at 10:41




Exactly! Thanks.
– Tegernako
Dec 9 '18 at 10:41











5














Now, use the fact that$$sum_{k=n+1}^mfrac1{k^2}<sum_{k=n+1}^mfrac1k-frac1{k-1}=frac1n-frac1m.$$






share|cite|improve this answer



















  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 '18 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 '18 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_{k=1}^infty frac{1}{k^2}$ is convergent, we know that that $sum_{k=n}^infty frac{1}{k^2}$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 '18 at 10:44








  • 1




    In order to prove that, for each $varepsilon>0$, $sum_{k=n+1}^mfrac1{k^2}<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 '18 at 10:46








  • 1




    @JonasLenz I guess that is how one shows that $sum_{k=1}^infty frac{1}{k^2}$ convergent.
    – Yanko
    Dec 9 '18 at 10:49
















5














Now, use the fact that$$sum_{k=n+1}^mfrac1{k^2}<sum_{k=n+1}^mfrac1k-frac1{k-1}=frac1n-frac1m.$$






share|cite|improve this answer



















  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 '18 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 '18 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_{k=1}^infty frac{1}{k^2}$ is convergent, we know that that $sum_{k=n}^infty frac{1}{k^2}$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 '18 at 10:44








  • 1




    In order to prove that, for each $varepsilon>0$, $sum_{k=n+1}^mfrac1{k^2}<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 '18 at 10:46








  • 1




    @JonasLenz I guess that is how one shows that $sum_{k=1}^infty frac{1}{k^2}$ convergent.
    – Yanko
    Dec 9 '18 at 10:49














5












5








5






Now, use the fact that$$sum_{k=n+1}^mfrac1{k^2}<sum_{k=n+1}^mfrac1k-frac1{k-1}=frac1n-frac1m.$$






share|cite|improve this answer














Now, use the fact that$$sum_{k=n+1}^mfrac1{k^2}<sum_{k=n+1}^mfrac1k-frac1{k-1}=frac1n-frac1m.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 10:39

























answered Dec 9 '18 at 10:36









José Carlos Santos

150k22121221




150k22121221








  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 '18 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 '18 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_{k=1}^infty frac{1}{k^2}$ is convergent, we know that that $sum_{k=n}^infty frac{1}{k^2}$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 '18 at 10:44








  • 1




    In order to prove that, for each $varepsilon>0$, $sum_{k=n+1}^mfrac1{k^2}<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 '18 at 10:46








  • 1




    @JonasLenz I guess that is how one shows that $sum_{k=1}^infty frac{1}{k^2}$ convergent.
    – Yanko
    Dec 9 '18 at 10:49














  • 1




    I've edited my answer. Thank you.
    – José Carlos Santos
    Dec 9 '18 at 10:39










  • Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
    – Tegernako
    Dec 9 '18 at 10:40






  • 1




    Why do we need the hint? Can't we simply say that as $sum_{k=1}^infty frac{1}{k^2}$ is convergent, we know that that $sum_{k=n}^infty frac{1}{k^2}$ converges to $0$ as $n to infty$ which shows the claim, too.
    – Jonas Lenz
    Dec 9 '18 at 10:44








  • 1




    In order to prove that, for each $varepsilon>0$, $sum_{k=n+1}^mfrac1{k^2}<varepsilon$, if $m$ and $n$ are large enough.
    – José Carlos Santos
    Dec 9 '18 at 10:46








  • 1




    @JonasLenz I guess that is how one shows that $sum_{k=1}^infty frac{1}{k^2}$ convergent.
    – Yanko
    Dec 9 '18 at 10:49








1




1




I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 '18 at 10:39




I've edited my answer. Thank you.
– José Carlos Santos
Dec 9 '18 at 10:39












Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 '18 at 10:40




Seems like I went into paralysis by analysis trying to over-express the sum I had. Thanks!
– Tegernako
Dec 9 '18 at 10:40




1




1




Why do we need the hint? Can't we simply say that as $sum_{k=1}^infty frac{1}{k^2}$ is convergent, we know that that $sum_{k=n}^infty frac{1}{k^2}$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 '18 at 10:44






Why do we need the hint? Can't we simply say that as $sum_{k=1}^infty frac{1}{k^2}$ is convergent, we know that that $sum_{k=n}^infty frac{1}{k^2}$ converges to $0$ as $n to infty$ which shows the claim, too.
– Jonas Lenz
Dec 9 '18 at 10:44






1




1




In order to prove that, for each $varepsilon>0$, $sum_{k=n+1}^mfrac1{k^2}<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 '18 at 10:46






In order to prove that, for each $varepsilon>0$, $sum_{k=n+1}^mfrac1{k^2}<varepsilon$, if $m$ and $n$ are large enough.
– José Carlos Santos
Dec 9 '18 at 10:46






1




1




@JonasLenz I guess that is how one shows that $sum_{k=1}^infty frac{1}{k^2}$ convergent.
– Yanko
Dec 9 '18 at 10:49




@JonasLenz I guess that is how one shows that $sum_{k=1}^infty frac{1}{k^2}$ convergent.
– Yanko
Dec 9 '18 at 10:49


















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