Show that sequence is a Cauchy sequence












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







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  • 1




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






  • 3




    $begingroup$
    Hint: a convergent sequence is Cauchy.
    $endgroup$
    – Bernard
    8 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
    8 hours ago






  • 3




    $begingroup$
    Hint: a convergent sequence is Cauchy.
    $endgroup$
    – Bernard
    8 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






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share|cite|improve this question













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edited 8 hours ago









Bernard

121k740116




121k740116










asked 8 hours ago









kira0705kira0705

1167




1167








  • 1




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






  • 3




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














  • 1




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






  • 3




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








1




1




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




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




3




3




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




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










3 Answers
3






active

oldest

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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.






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  • 1




    $begingroup$
    Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
    $endgroup$
    – kira0705
    7 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









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    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











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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

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      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
        7 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
        7 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 8 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
        7 hours ago














      • 1




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








      1




      1




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




      $begingroup$
      Thank you for such an elaborate proof , not ignoring the signs was an important step indeed.
      $endgroup$
      – kira0705
      7 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 8 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 $$|f_n - f_{n+1}| = |frac{1}{n} + frac{1}{n+1}| = |frac{n+2}{n(n+1)}| = frac{n+2}{n(n+1)} < frac{n+2}{n^2} =frac{1}{n} +frac{2}{n^2} < frac{1}{frac{2}{sqrt{epsilon}}} + frac{epsilon}{2} = frac{sqrt{epsilon}}{2} + frac{epsilon}{2} < frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$






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              $endgroup$


















                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 7 hours ago

























                  answered 8 hours ago









                  user516079user516079

                  318210




                  318210






























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