Show that sequence is a Cauchy sequenceLet $a_n=frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $ngeq 2$....

What is an explicit bijection in combinatorics?

What does "south of due west" mean?

Algebraic proof that two statements of the fundamental theorem of algebra are equivalent

Is the UK legally prevented from having another referendum on Brexit?

How do I fight with Heavy Armor as a Wizard with Tenser's Transformation?

How can guns be countered by melee combat without raw-ability or exceptional explanations?

Isn't a semicolon (';') needed after a function declaration in C++?

Can you say "leftside right"?

How can I give a Ranger advantage on a check due to Favored Enemy without spoiling the story for the player?

Taking an academic pseudonym?

Why don't programs completely uninstall (remove all their files) when I remove them?

When distributing a Linux kernel driver as source code, what's the difference between Proprietary and GPL license?

Cartoon in which kids compete by fighting as monsters or creatures with special powers in a virtual world

Missing a connection and don't have money to book next flight

Partial derivative with respect to three variables

Piping Multiple Numbers into Sed

What sort of grammatical construct is ‘Quod per sortem sternit fortem’?

Boss asked me to sign a resignation paper without a date on it along with my new contract

bash aliases do not expand even with shopt expand_aliases

Why is it that Bernie Sanders always called a "socialist"?

How many copper coins fit inside a cubic foot?

Short story about a man betting a group he could tell a story, and one of them would disappear and the others would not notice

What is formjacking?

Are one-line email responses considered disrespectful?



Show that sequence is a Cauchy sequence


Let $a_n=frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $ngeq 2$. Show that ${a_n}_{n=1}^{infty}$ is CauchyShow that $langle f_n rangle$ is a Cauchy sequence, where $f_n=1-frac12+frac13-frac14+dots+frac{(-1)^{n-1}}{n}$Show that $(3x_{n}+4y_{n})$ is also Cauchy sequence.Uniformly Cauchy sequence of functionsIs this sequence Cauchy in the space of polynomials under the infinity norm?The sequence $b_n=pa_n+qa_{n+1}$, where $|p|<q$, is convergent. Prove that $a_n$ converges. If $|p|geq q > 0$ show that $a_n$ need not converge.Prove that it is a cauchy sequenceShowing sequence is Cauchy by DefinitionSequence of function on $mathbb{R}$ Cauchy iff convergentProve that $(a_n) preccurlyeq_1 (b_n) iff (a_n) preccurlyeq_2 (b_n)$ or $(a_n) approx (b_n)$ for Cauchy sequences













2












$begingroup$


Prove that given sequence $$langle f_nrangle =1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{(-1)^{n-1}}{n}$$



is a Cauchy sequence



My attempt :
$|f_{n}-f_{m}|=Biggl|dfrac{(-1)^{m}}{m+1}+dfrac{(-1)^{m+1}}{m+2}cdotsdots+dfrac{(-1)^{n-1}}{n}Biggr|$



using $ m+1>m implies dfrac{1}{m+1}<dfrac{1}{m} $



$|f_{n}-f_{m}|le dfrac{1}{m}+dfrac{1}{m}+dfrac{1}{m}cdotscdotsdfrac{1}{m}$



$|f_{n}-f_{m}|ledfrac{n-m}{m}$



I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .






sequences-and-series cauchy-sequences







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, the limit of the sequence because of Leibniz' criterion.
    $endgroup$
    – egreg
    4 hours ago






  • 1




    $begingroup$
    Hint: a convergent sequence is Cauchy.
    $endgroup$
    – Bernard
    4 hours ago
















2












$begingroup$


Prove that given sequence $$langle f_nrangle =1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{(-1)^{n-1}}{n}$$



is a Cauchy sequence



My attempt :
$|f_{n}-f_{m}|=Biggl|dfrac{(-1)^{m}}{m+1}+dfrac{(-1)^{m+1}}{m+2}cdotsdots+dfrac{(-1)^{n-1}}{n}Biggr|$



using $ m+1>m implies dfrac{1}{m+1}<dfrac{1}{m} $



$|f_{n}-f_{m}|le dfrac{1}{m}+dfrac{1}{m}+dfrac{1}{m}cdotscdotsdfrac{1}{m}$



$|f_{n}-f_{m}|ledfrac{n-m}{m}$



I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .






sequences-and-series cauchy-sequences







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, the limit of the sequence because of Leibniz' criterion.
    $endgroup$
    – egreg
    4 hours ago






  • 1




    $begingroup$
    Hint: a convergent sequence is Cauchy.
    $endgroup$
    – Bernard
    4 hours ago














2












2








2





$begingroup$


Prove that given sequence $$langle f_nrangle =1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{(-1)^{n-1}}{n}$$



is a Cauchy sequence



My attempt :
$|f_{n}-f_{m}|=Biggl|dfrac{(-1)^{m}}{m+1}+dfrac{(-1)^{m+1}}{m+2}cdotsdots+dfrac{(-1)^{n-1}}{n}Biggr|$



using $ m+1>m implies dfrac{1}{m+1}<dfrac{1}{m} $



$|f_{n}-f_{m}|le dfrac{1}{m}+dfrac{1}{m}+dfrac{1}{m}cdotscdotsdfrac{1}{m}$



$|f_{n}-f_{m}|ledfrac{n-m}{m}$



I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .






sequences-and-series cauchy-sequences







share|cite|improve this question











$endgroup$




Prove that given sequence $$langle f_nrangle =1-frac{1}{2}+frac{1}{3}-frac{1}{4}+.....+frac{(-1)^{n-1}}{n}$$



is a Cauchy sequence



My attempt :
$|f_{n}-f_{m}|=Biggl|dfrac{(-1)^{m}}{m+1}+dfrac{(-1)^{m+1}}{m+2}cdotsdots+dfrac{(-1)^{n-1}}{n}Biggr|$



using $ m+1>m implies dfrac{1}{m+1}<dfrac{1}{m} $



$|f_{n}-f_{m}|le dfrac{1}{m}+dfrac{1}{m}+dfrac{1}{m}cdotscdotsdfrac{1}{m}$



$|f_{n}-f_{m}|ledfrac{n-m}{m}$



I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .






sequences-and-series cauchy-sequences




sequences-and-series cauchy-sequences






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Bernard

121k740116




121k740116










asked 4 hours ago









kira0705kira0705

1167




1167








  • 1




    $begingroup$
    Well, the limit of the sequence because of Leibniz' criterion.
    $endgroup$
    – egreg
    4 hours ago






  • 1




    $begingroup$
    Hint: a convergent sequence is Cauchy.
    $endgroup$
    – Bernard
    4 hours ago














  • 1




    $begingroup$
    Well, the limit of the sequence because of Leibniz' criterion.
    $endgroup$
    – egreg
    4 hours ago






  • 1




    $begingroup$
    Hint: a convergent sequence is Cauchy.
    $endgroup$
    – Bernard
    4 hours ago








1




1




$begingroup$
Well, the limit of the sequence because of Leibniz' criterion.
$endgroup$
– egreg
4 hours ago




$begingroup$
Well, the limit of the sequence because of Leibniz' criterion.
$endgroup$
– egreg
4 hours ago




1




1




$begingroup$
Hint: a convergent sequence is Cauchy.
$endgroup$
– Bernard
4 hours ago




$begingroup$
Hint: a convergent sequence is Cauchy.
$endgroup$
– Bernard
4 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

If you ignore the signs of the terms,
the result diverges.
So you can't do that.



$f_n
=sum_{k=1}^n dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=sum_{k=m+1}^n dfrac{(-1)^k}{k}
=sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
=(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
$.



If
$n-m$ is even,
so $n-m = 2j$,
then



$begin{array}\
f_n-f_m
&=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
&=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
&=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
&=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
text{so}\
|f_n-f_m|
&=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
&=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
quadtext{this is the sneaky part}\
&lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
&= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
&= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
&< dfrac{1}{2m}\
&to 0 text{ as } m to infty\
end{array}
$



If $n-m$ is odd,
the sum changes
by at most $frac1{n}$
so it still goes to zero.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
    $endgroup$
    – kira0705
    3 hours ago



















2












$begingroup$

Hint :



$$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$






    share|cite|improve this answer











    $endgroup$













      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%2f3123120%2fshow-that-sequence-is-a-cauchy-sequence%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      If you ignore the signs of the terms,
      the result diverges.
      So you can't do that.



      $f_n
      =sum_{k=1}^n dfrac{(-1)^k}{k}
      $
      so,
      if $n > m$,
      $f_n-f_m
      =sum_{k=m+1}^n dfrac{(-1)^k}{k}
      =sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
      =(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
      $      .



      If
      $n-m$ is even,
      so $n-m = 2j$,
      then



      $begin{array}\
      f_n-f_m
      &=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
      &=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      text{so}\
      |f_n-f_m|
      &=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      &=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
      quadtext{this is the sneaky part}\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
      &= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
      &< dfrac{1}{2m}\
      &to 0 text{ as } m to infty\
      end{array}
      $



      If $n-m$ is odd,
      the sum changes
      by at most $frac1{n}$
      so it still goes to zero.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
        $endgroup$
        – kira0705
        3 hours ago
















      2












      $begingroup$

      If you ignore the signs of the terms,
      the result diverges.
      So you can't do that.



      $f_n
      =sum_{k=1}^n dfrac{(-1)^k}{k}
      $
      so,
      if $n > m$,
      $f_n-f_m
      =sum_{k=m+1}^n dfrac{(-1)^k}{k}
      =sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
      =(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
      $      .



      If
      $n-m$ is even,
      so $n-m = 2j$,
      then



      $begin{array}\
      f_n-f_m
      &=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
      &=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      text{so}\
      |f_n-f_m|
      &=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      &=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
      quadtext{this is the sneaky part}\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
      &= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
      &< dfrac{1}{2m}\
      &to 0 text{ as } m to infty\
      end{array}
      $



      If $n-m$ is odd,
      the sum changes
      by at most $frac1{n}$
      so it still goes to zero.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
        $endgroup$
        – kira0705
        3 hours ago














      2












      2








      2





      $begingroup$

      If you ignore the signs of the terms,
      the result diverges.
      So you can't do that.



      $f_n
      =sum_{k=1}^n dfrac{(-1)^k}{k}
      $
      so,
      if $n > m$,
      $f_n-f_m
      =sum_{k=m+1}^n dfrac{(-1)^k}{k}
      =sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
      =(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
      $      .



      If
      $n-m$ is even,
      so $n-m = 2j$,
      then



      $begin{array}\
      f_n-f_m
      &=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
      &=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      text{so}\
      |f_n-f_m|
      &=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      &=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
      quadtext{this is the sneaky part}\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
      &= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
      &< dfrac{1}{2m}\
      &to 0 text{ as } m to infty\
      end{array}
      $



      If $n-m$ is odd,
      the sum changes
      by at most $frac1{n}$
      so it still goes to zero.






      share|cite|improve this answer









      $endgroup$



      If you ignore the signs of the terms,
      the result diverges.
      So you can't do that.



      $f_n
      =sum_{k=1}^n dfrac{(-1)^k}{k}
      $
      so,
      if $n > m$,
      $f_n-f_m
      =sum_{k=m+1}^n dfrac{(-1)^k}{k}
      =sum_{k=1}^{n-m} dfrac{(-1)^{k+m}}{k+m}
      =(-1)^msum_{k=1}^{n-m} dfrac{(-1)^{k}}{k+m}
      $      .



      If
      $n-m$ is even,
      so $n-m = 2j$,
      then



      $begin{array}\
      f_n-f_m
      &=(-1)^msum_{k=1}^{2j} dfrac{(-1)^{k}}{k+m}\
      &=(-1)^msum_{k=1}^{j} left(dfrac{(-1)^{2k-1}}{2k-1+m}+dfrac{(-1)^{2k}}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{-1}{2k-1+m}+dfrac{1}{2k+m}right)\
      &=(-1)^msum_{k=1}^{j} (-1)^{2k-1}left(dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m+1}sum_{k=1}^{j} left(dfrac{-1}{(2k-1+m)(2k+m)}right)\
      &=(-1)^{m}sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      text{so}\
      |f_n-f_m|
      &=sum_{k=1}^{j} left(dfrac{1}{(2k-1+m)(2k+m)}right)\
      &=sum_{k=1}^{j}dfrac14 left(dfrac{1}{(k-frac12+frac{m}{2})(k+frac{m}{2})}right)\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{(k-1+frac{m}{2})(k+frac{m}{2})}right)
      quadtext{this is the sneaky part}\
      &lt dfrac14sum_{k=1}^{j} left(dfrac{1}{k-1+frac{m}{2}}-dfrac{1}{k+frac{m}{2}}right)\
      &= dfrac14 left(dfrac{1}{frac{m}{2}}-dfrac{1}{j+frac{m}{2}}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{2j+m}right)\
      &= dfrac12 left(dfrac{1}{m}-dfrac{1}{n}right)\
      &< dfrac{1}{2m}\
      &to 0 text{ as } m to infty\
      end{array}
      $



      If $n-m$ is odd,
      the sum changes
      by at most $frac1{n}$
      so it still goes to zero.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 3 hours ago









      marty cohenmarty cohen

      73.8k549128




      73.8k549128








      • 1




        $begingroup$
        Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
        $endgroup$
        – kira0705
        3 hours ago














      • 1




        $begingroup$
        Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
        $endgroup$
        – kira0705
        3 hours ago








      1




      1




      $begingroup$
      Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
      $endgroup$
      – kira0705
      3 hours ago




      $begingroup$
      Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
      $endgroup$
      – kira0705
      3 hours ago











      2












      $begingroup$

      Hint :



      $$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Hint :



        $$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Hint :



          $$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$






          share|cite|improve this answer









          $endgroup$



          Hint :



          $$frac{1}{2n-1}-frac{1}{2n}=frac{1}{(2n-1)2n}leq frac{1}{(n-1)n}= frac{1}{n-1}-frac{1}{n} $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Clément GuérinClément Guérin

          10k1736




          10k1736























              0












              $begingroup$

              Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$






                  share|cite|improve this answer











                  $endgroup$



                  Fix $epsilon > 0$, then for $n > frac{2}{sqrt{epsilon}}$ (So since both sides of the inequality are positive, $n^2 > frac{4}{epsilon} implies frac{epsilon}{2} > frac{2}{n^2}$), observe that $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 3 hours ago

























                  answered 3 hours ago









                  user516079user516079

                  318210




                  318210






























                      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.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3123120%2fshow-that-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

                      Щит и меч (фильм) Содержание Названия серий | Сюжет |...

                      Image
                      Фильмы по алфавитуФильмы 1968 годаФильмы СССР 1968 годаЧёрно-белые фильмы СССРФильмы студии «Мосфильм»Лидеры советского кинопрокатаФильмы-драмы СССРФильмы о Второй мировой войнеФильмы о шпионахЭкранизации произведений Вадима КожевниковаФильмы Владимира БасоваФильмы о фашизме 1968 годуодноимённому романуВадима КожевниковаВладимиром БасовымСтанислав ЛюбшинГерманиюАбверегауптштурмфюрераССБерлинслужбу СДSchwarzbraun ist die HaselnussXVIII векаЭрикиХермса НиляГитлерюгендаАбвераАх, мой милый АвгустинXVII векаHorst WesselDeutschland über alles У этого термина существуют и другие значения, см. Щит и меч. Щит и меч Жанр драма военный фильм Режиссёр Владимир Басов Автор сценария Владимир Басов Вадим Кожевников В главных ролях Станислав Любшин Валентина Титова Олег Янковский Георгий Мартынюк Владимир Басов Оператор Тимофей Лебешев Композитор Вениамин Баснер Кинокомпания Киностудия «Мосфильм». Творческое объединение писателей и киноработников
                      Read more

                      is 'sed' thread safeWhat should someone know about using Python scripts in the shell?Nexenta bash script uses...

                      Image
                      PTIJ: Aharon, King of Egypt Can a space-faring robot still function over a billion years? The need of reserving one's ability in job interviews Plagiarism of code by other PhD student Reason why dimensional travelling would be restricted Book about a time-travel war fought by computers It doesn't matter the side you see it What is a term for a function that when called repeatedly, has the same effect as calling once? Why doesn't "adolescent" take any articles in "listen to adolescent agonising"? Why would the IRS ask for birth certificates or even audit a small tax return?
                      Read more

                      Meter-Bus Содержание Параметры шины | Стандартизация |...

                      Image
                      RS-232IrDAHARTмодемтоковая петляИРПСMIDIIO-LinkRS-422RS-423RS-485LINM-BusKansas City StandardКомпакт-кассетагрампластинкаPPPSLIPIrDAHARTISO 7816MIDIIO-LinkModbusLINDMX-512P-NETProfibusIEC 608708250 UART16550 UART Сетевые протоколыПоследовательные шины физического уровняполевой шиныасинхронного интерфейсаOpen Metering SystemAESздесь У этого термина существуют и другие значения, см. M-bus (значения). M-Bus (Meter-Bus) — стандарт физического уровня для полевой шины на основе асинхронного интерфейса. Также под этим названием понимают коммуникационный протокол, используемый для связи устройств по этой шине. M-bus преимущественно применяется для приборов учета электрической энергии (электросчётчики), тепловой энергии (теплосчётчики), расходомеров воды и газа. Данные передаются на компьютерную станцию (сервер) напрямую или через концентраторы шины M-Bus, а также усилители-повторители сигнала. Содержание 1 Параметры шины 2 Стандартизация 2.1 Open Metering System
                      Read more