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 at 10:50










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










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









Jonas Lenz

530212




530212










asked Dec 9 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 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


















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 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 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 $sum_{k=1}^infty frac{1}{k^2}$ 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 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 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 at 10:49











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032238%2fdetermine-if-this-specific-sequence-is-a-cauchy-sequence%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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 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 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 at 10:51

























answered Dec 9 at 10:35









Yanko

6,029724




6,029724












  • 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











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 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 $sum_{k=1}^infty frac{1}{k^2}$ 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 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 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 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 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 $sum_{k=1}^infty frac{1}{k^2}$ 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 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 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 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 at 10:39

























answered Dec 9 at 10:36









José Carlos Santos

150k22119221




150k22119221








  • 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 $sum_{k=1}^infty frac{1}{k^2}$ 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 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 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 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 $sum_{k=1}^infty frac{1}{k^2}$ 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 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 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 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 $sum_{k=1}^infty frac{1}{k^2}$ 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 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 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 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 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 at 10:49




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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032238%2fdetermine-if-this-specific-sequence-is-a-cauchy-sequence%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more