Prove $sum_{k=1}^nfrac{1}{n+k} geq frac{7}{12}$












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$begingroup$


For all integers $ngeq2$,



$$sum_{k=1}^nfrac{1}{n+k} geq frac{7}{12}$$



How would you prove this by induction?






inequality proof-explanation harmonic-numbers







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$endgroup$












  • $begingroup$
    $frac 1 {n+1+k} < frac 1 {n+k} $. Do you know by how much. Can you prove $sum (frac 1 {n+1+k} - frac 1 {n+k} + frac 1 {2 (n+1)} le 0$?
    $endgroup$
    – fleablood
    Jan 26 '17 at 1:46


















0












$begingroup$


For all integers $ngeq2$,



$$sum_{k=1}^nfrac{1}{n+k} geq frac{7}{12}$$



How would you prove this by induction?






inequality proof-explanation harmonic-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    $frac 1 {n+1+k} < frac 1 {n+k} $. Do you know by how much. Can you prove $sum (frac 1 {n+1+k} - frac 1 {n+k} + frac 1 {2 (n+1)} le 0$?
    $endgroup$
    – fleablood
    Jan 26 '17 at 1:46
















0












0








0





$begingroup$


For all integers $ngeq2$,



$$sum_{k=1}^nfrac{1}{n+k} geq frac{7}{12}$$



How would you prove this by induction?






inequality proof-explanation harmonic-numbers







share|cite|improve this question











$endgroup$




For all integers $ngeq2$,



$$sum_{k=1}^nfrac{1}{n+k} geq frac{7}{12}$$



How would you prove this by induction?






inequality proof-explanation harmonic-numbers




inequality proof-explanation harmonic-numbers






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 31 '18 at 9:45









Martin Sleziak

44.7k9117272




44.7k9117272










asked Jan 26 '17 at 1:34









IvyIvy

95112




95112












  • $begingroup$
    $frac 1 {n+1+k} < frac 1 {n+k} $. Do you know by how much. Can you prove $sum (frac 1 {n+1+k} - frac 1 {n+k} + frac 1 {2 (n+1)} le 0$?
    $endgroup$
    – fleablood
    Jan 26 '17 at 1:46




















  • $begingroup$
    $frac 1 {n+1+k} < frac 1 {n+k} $. Do you know by how much. Can you prove $sum (frac 1 {n+1+k} - frac 1 {n+k} + frac 1 {2 (n+1)} le 0$?
    $endgroup$
    – fleablood
    Jan 26 '17 at 1:46


















$begingroup$
$frac 1 {n+1+k} < frac 1 {n+k} $. Do you know by how much. Can you prove $sum (frac 1 {n+1+k} - frac 1 {n+k} + frac 1 {2 (n+1)} le 0$?
$endgroup$
– fleablood
Jan 26 '17 at 1:46






$begingroup$
$frac 1 {n+1+k} < frac 1 {n+k} $. Do you know by how much. Can you prove $sum (frac 1 {n+1+k} - frac 1 {n+k} + frac 1 {2 (n+1)} le 0$?
$endgroup$
– fleablood
Jan 26 '17 at 1:46












3 Answers
3






active

oldest

votes


















3












$begingroup$

$$sum_{k=1}^{n}frac{1}{n+k} = H_{2n}-H_{n}$$
are the terms of an increasing sequence, since
$$ H_{2n+2}-H_{n+1}-H_{2n}+H_n = frac{1}{2n+2}+frac{1}{2n+1}-frac{1}{n+1} = frac{1}{(2n+1)(2n+2)}>0.$$






share|cite|improve this answer









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  • 1




    $begingroup$
    I don't see how this helps though?
    $endgroup$
    – Ivy
    Jan 26 '17 at 1:44










  • $begingroup$
    @Ivy: $sum_{k=1}^{2}frac{1}{2+k}=frac{7}{12}$ and I have just proved the next terms are bigger.
    $endgroup$
    – Jack D'Aurizio
    Jan 26 '17 at 1:46



















0












$begingroup$

By induction you can do:



$$sum_{k=1}^{n+1}frac{1}{n+1+k}=sum_{k=1}^nfrac{1}{n+k} -frac{1}{n+1}+frac{1}{2n+1}+frac{1}{2n+2}=\
=sum_{k=1}^nfrac{1}{n+k}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    If
    $s_n
    = sum_{k=1}^nfrac{1}{n+k}
    $,
    then



    $begin{array}\
    s_{n+1}
    &= sum_{k=1}^{n+1}dfrac{1}{n+1+k}\
    &= sum_{k=2}^{n+2}dfrac{1}{n+k}\
    &= sum_{k=2}^{n}dfrac{1}{n+k}+dfrac1{2n+1}+dfrac1{2n+2}\
    &= sum_{k=1}^{n}dfrac{1}{n+k}-dfrac1{n+1}+dfrac1{2n+1}+dfrac1{2n+2}\
    &= s_n+dfrac{(2n+1)+(2n+2)-2(2n+1)}{(2n+1)(2n+2)}\
    &= s_n+dfrac{1}{(2n+1)(2n+2)}\
    end{array}
    $



    Therefore
    $s_{n+1} > s_n$,
    so $s_n$
    is increasing.



    Therefore,
    if $n > m$,
    $s_n > s_m$.



    Note that
    we can bound $s_n$ because



    $begin{array}\
    s_{n+1}
    &= s_n+dfrac{1}{(2n+1)(2n+2)}\
    &< s_n+dfrac{1}{(2n)(2n+2)}\
    &< s_n+frac14dfrac{1}{n(n+1)}\
    &< s_n+frac14(dfrac1{n}-dfrac1{n+1})\
    end{array}
    $



    so that
    $s_{n+1}-s_n
    < frac14(dfrac1{n}-dfrac1{n+1})
    $.



    Summing,
    $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
    < sum_{k=1}^{m-1}frac14(dfrac1{n}-dfrac1{n+1})
    $
    or
    $s_{n+m}-s_n
    lt frac14(frac1{n}-frac1{n+m})
    lt frac1{4n}
    $.



    Therefore,
    $lim_{n to infty} s_n$
    exists,
    and,
    if
    $S=lim_{n to infty} s_n$,
    $S < s_n +dfrac1{4n}$.



    We can similarly get a
    lower bound on $s_n$
    and, therefore,
    a lower bound on $S$
    in terms of $n$ and $s_n$.



    (the following is done
    with copy, paste, and edit.)



    $begin{array}\
    s_{n+1}
    &= s_n+dfrac{1}{(2n+1)(2n+2)}\
    &> s_n+dfrac{1}{(2n+2)(2n+4)}\
    &> s_n+frac14dfrac{1}{(n+2)(n+2)}\
    &> s_n+frac14(dfrac1{n+1}-dfrac1{n+2})\
    end{array}
    $



    so that
    $s_{n+1}-s_n
    gt frac14(dfrac1{n+1}-dfrac1{n+2})
    $.



    Summing,
    $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
    gt sum_{k=1}^{m-1}frac14(dfrac1{n+1}-dfrac1{n+2})
    $
    or
    $s_{n+m}-s_n
    gt frac14(frac1{n+1}-frac1{n+m+1})
    $
    so that
    $S
    =lim_{m to infty}s_{n+m}
    ge s_n+lim_{m to infty}frac14(frac1{n+1}-frac1{n+m+1})
    = s_n+frac1{4n+4}
    $.



    Therefore
    $S ge s_n +dfrac1{4n+4}$.



    Therefore,
    for any $n$,
    $s_n +dfrac1{4n+4}
    le S
    < s_n +dfrac1{4n}$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      3












      $begingroup$

      $$sum_{k=1}^{n}frac{1}{n+k} = H_{2n}-H_{n}$$
      are the terms of an increasing sequence, since
      $$ H_{2n+2}-H_{n+1}-H_{2n}+H_n = frac{1}{2n+2}+frac{1}{2n+1}-frac{1}{n+1} = frac{1}{(2n+1)(2n+2)}>0.$$






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        I don't see how this helps though?
        $endgroup$
        – Ivy
        Jan 26 '17 at 1:44










      • $begingroup$
        @Ivy: $sum_{k=1}^{2}frac{1}{2+k}=frac{7}{12}$ and I have just proved the next terms are bigger.
        $endgroup$
        – Jack D'Aurizio
        Jan 26 '17 at 1:46
















      3












      $begingroup$

      $$sum_{k=1}^{n}frac{1}{n+k} = H_{2n}-H_{n}$$
      are the terms of an increasing sequence, since
      $$ H_{2n+2}-H_{n+1}-H_{2n}+H_n = frac{1}{2n+2}+frac{1}{2n+1}-frac{1}{n+1} = frac{1}{(2n+1)(2n+2)}>0.$$






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        I don't see how this helps though?
        $endgroup$
        – Ivy
        Jan 26 '17 at 1:44










      • $begingroup$
        @Ivy: $sum_{k=1}^{2}frac{1}{2+k}=frac{7}{12}$ and I have just proved the next terms are bigger.
        $endgroup$
        – Jack D'Aurizio
        Jan 26 '17 at 1:46














      3












      3








      3





      $begingroup$

      $$sum_{k=1}^{n}frac{1}{n+k} = H_{2n}-H_{n}$$
      are the terms of an increasing sequence, since
      $$ H_{2n+2}-H_{n+1}-H_{2n}+H_n = frac{1}{2n+2}+frac{1}{2n+1}-frac{1}{n+1} = frac{1}{(2n+1)(2n+2)}>0.$$






      share|cite|improve this answer









      $endgroup$



      $$sum_{k=1}^{n}frac{1}{n+k} = H_{2n}-H_{n}$$
      are the terms of an increasing sequence, since
      $$ H_{2n+2}-H_{n+1}-H_{2n}+H_n = frac{1}{2n+2}+frac{1}{2n+1}-frac{1}{n+1} = frac{1}{(2n+1)(2n+2)}>0.$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 26 '17 at 1:39









      Jack D'AurizioJack D'Aurizio

      288k33280660




      288k33280660








      • 1




        $begingroup$
        I don't see how this helps though?
        $endgroup$
        – Ivy
        Jan 26 '17 at 1:44










      • $begingroup$
        @Ivy: $sum_{k=1}^{2}frac{1}{2+k}=frac{7}{12}$ and I have just proved the next terms are bigger.
        $endgroup$
        – Jack D'Aurizio
        Jan 26 '17 at 1:46














      • 1




        $begingroup$
        I don't see how this helps though?
        $endgroup$
        – Ivy
        Jan 26 '17 at 1:44










      • $begingroup$
        @Ivy: $sum_{k=1}^{2}frac{1}{2+k}=frac{7}{12}$ and I have just proved the next terms are bigger.
        $endgroup$
        – Jack D'Aurizio
        Jan 26 '17 at 1:46








      1




      1




      $begingroup$
      I don't see how this helps though?
      $endgroup$
      – Ivy
      Jan 26 '17 at 1:44




      $begingroup$
      I don't see how this helps though?
      $endgroup$
      – Ivy
      Jan 26 '17 at 1:44












      $begingroup$
      @Ivy: $sum_{k=1}^{2}frac{1}{2+k}=frac{7}{12}$ and I have just proved the next terms are bigger.
      $endgroup$
      – Jack D'Aurizio
      Jan 26 '17 at 1:46




      $begingroup$
      @Ivy: $sum_{k=1}^{2}frac{1}{2+k}=frac{7}{12}$ and I have just proved the next terms are bigger.
      $endgroup$
      – Jack D'Aurizio
      Jan 26 '17 at 1:46











      0












      $begingroup$

      By induction you can do:



      $$sum_{k=1}^{n+1}frac{1}{n+1+k}=sum_{k=1}^nfrac{1}{n+k} -frac{1}{n+1}+frac{1}{2n+1}+frac{1}{2n+2}=\
      =sum_{k=1}^nfrac{1}{n+k}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}$$






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        By induction you can do:



        $$sum_{k=1}^{n+1}frac{1}{n+1+k}=sum_{k=1}^nfrac{1}{n+k} -frac{1}{n+1}+frac{1}{2n+1}+frac{1}{2n+2}=\
        =sum_{k=1}^nfrac{1}{n+k}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}$$






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          By induction you can do:



          $$sum_{k=1}^{n+1}frac{1}{n+1+k}=sum_{k=1}^nfrac{1}{n+k} -frac{1}{n+1}+frac{1}{2n+1}+frac{1}{2n+2}=\
          =sum_{k=1}^nfrac{1}{n+k}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}$$






          share|cite|improve this answer











          $endgroup$



          By induction you can do:



          $$sum_{k=1}^{n+1}frac{1}{n+1+k}=sum_{k=1}^nfrac{1}{n+k} -frac{1}{n+1}+frac{1}{2n+1}+frac{1}{2n+2}=\
          =sum_{k=1}^nfrac{1}{n+k}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}+frac{1}{(2n+1)(2n+2)}ge frac{7}{12}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 26 '17 at 20:04

























          answered Jan 26 '17 at 2:01









          ArnaldoArnaldo

          18.1k42246




          18.1k42246























              0












              $begingroup$

              If
              $s_n
              = sum_{k=1}^nfrac{1}{n+k}
              $,
              then



              $begin{array}\
              s_{n+1}
              &= sum_{k=1}^{n+1}dfrac{1}{n+1+k}\
              &= sum_{k=2}^{n+2}dfrac{1}{n+k}\
              &= sum_{k=2}^{n}dfrac{1}{n+k}+dfrac1{2n+1}+dfrac1{2n+2}\
              &= sum_{k=1}^{n}dfrac{1}{n+k}-dfrac1{n+1}+dfrac1{2n+1}+dfrac1{2n+2}\
              &= s_n+dfrac{(2n+1)+(2n+2)-2(2n+1)}{(2n+1)(2n+2)}\
              &= s_n+dfrac{1}{(2n+1)(2n+2)}\
              end{array}
              $



              Therefore
              $s_{n+1} > s_n$,
              so $s_n$
              is increasing.



              Therefore,
              if $n > m$,
              $s_n > s_m$.



              Note that
              we can bound $s_n$ because



              $begin{array}\
              s_{n+1}
              &= s_n+dfrac{1}{(2n+1)(2n+2)}\
              &< s_n+dfrac{1}{(2n)(2n+2)}\
              &< s_n+frac14dfrac{1}{n(n+1)}\
              &< s_n+frac14(dfrac1{n}-dfrac1{n+1})\
              end{array}
              $



              so that
              $s_{n+1}-s_n
              < frac14(dfrac1{n}-dfrac1{n+1})
              $.



              Summing,
              $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
              < sum_{k=1}^{m-1}frac14(dfrac1{n}-dfrac1{n+1})
              $
              or
              $s_{n+m}-s_n
              lt frac14(frac1{n}-frac1{n+m})
              lt frac1{4n}
              $.



              Therefore,
              $lim_{n to infty} s_n$
              exists,
              and,
              if
              $S=lim_{n to infty} s_n$,
              $S < s_n +dfrac1{4n}$.



              We can similarly get a
              lower bound on $s_n$
              and, therefore,
              a lower bound on $S$
              in terms of $n$ and $s_n$.



              (the following is done
              with copy, paste, and edit.)



              $begin{array}\
              s_{n+1}
              &= s_n+dfrac{1}{(2n+1)(2n+2)}\
              &> s_n+dfrac{1}{(2n+2)(2n+4)}\
              &> s_n+frac14dfrac{1}{(n+2)(n+2)}\
              &> s_n+frac14(dfrac1{n+1}-dfrac1{n+2})\
              end{array}
              $



              so that
              $s_{n+1}-s_n
              gt frac14(dfrac1{n+1}-dfrac1{n+2})
              $.



              Summing,
              $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
              gt sum_{k=1}^{m-1}frac14(dfrac1{n+1}-dfrac1{n+2})
              $
              or
              $s_{n+m}-s_n
              gt frac14(frac1{n+1}-frac1{n+m+1})
              $
              so that
              $S
              =lim_{m to infty}s_{n+m}
              ge s_n+lim_{m to infty}frac14(frac1{n+1}-frac1{n+m+1})
              = s_n+frac1{4n+4}
              $.



              Therefore
              $S ge s_n +dfrac1{4n+4}$.



              Therefore,
              for any $n$,
              $s_n +dfrac1{4n+4}
              le S
              < s_n +dfrac1{4n}$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If
                $s_n
                = sum_{k=1}^nfrac{1}{n+k}
                $,
                then



                $begin{array}\
                s_{n+1}
                &= sum_{k=1}^{n+1}dfrac{1}{n+1+k}\
                &= sum_{k=2}^{n+2}dfrac{1}{n+k}\
                &= sum_{k=2}^{n}dfrac{1}{n+k}+dfrac1{2n+1}+dfrac1{2n+2}\
                &= sum_{k=1}^{n}dfrac{1}{n+k}-dfrac1{n+1}+dfrac1{2n+1}+dfrac1{2n+2}\
                &= s_n+dfrac{(2n+1)+(2n+2)-2(2n+1)}{(2n+1)(2n+2)}\
                &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                end{array}
                $



                Therefore
                $s_{n+1} > s_n$,
                so $s_n$
                is increasing.



                Therefore,
                if $n > m$,
                $s_n > s_m$.



                Note that
                we can bound $s_n$ because



                $begin{array}\
                s_{n+1}
                &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                &< s_n+dfrac{1}{(2n)(2n+2)}\
                &< s_n+frac14dfrac{1}{n(n+1)}\
                &< s_n+frac14(dfrac1{n}-dfrac1{n+1})\
                end{array}
                $



                so that
                $s_{n+1}-s_n
                < frac14(dfrac1{n}-dfrac1{n+1})
                $.



                Summing,
                $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
                < sum_{k=1}^{m-1}frac14(dfrac1{n}-dfrac1{n+1})
                $
                or
                $s_{n+m}-s_n
                lt frac14(frac1{n}-frac1{n+m})
                lt frac1{4n}
                $.



                Therefore,
                $lim_{n to infty} s_n$
                exists,
                and,
                if
                $S=lim_{n to infty} s_n$,
                $S < s_n +dfrac1{4n}$.



                We can similarly get a
                lower bound on $s_n$
                and, therefore,
                a lower bound on $S$
                in terms of $n$ and $s_n$.



                (the following is done
                with copy, paste, and edit.)



                $begin{array}\
                s_{n+1}
                &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                &> s_n+dfrac{1}{(2n+2)(2n+4)}\
                &> s_n+frac14dfrac{1}{(n+2)(n+2)}\
                &> s_n+frac14(dfrac1{n+1}-dfrac1{n+2})\
                end{array}
                $



                so that
                $s_{n+1}-s_n
                gt frac14(dfrac1{n+1}-dfrac1{n+2})
                $.



                Summing,
                $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
                gt sum_{k=1}^{m-1}frac14(dfrac1{n+1}-dfrac1{n+2})
                $
                or
                $s_{n+m}-s_n
                gt frac14(frac1{n+1}-frac1{n+m+1})
                $
                so that
                $S
                =lim_{m to infty}s_{n+m}
                ge s_n+lim_{m to infty}frac14(frac1{n+1}-frac1{n+m+1})
                = s_n+frac1{4n+4}
                $.



                Therefore
                $S ge s_n +dfrac1{4n+4}$.



                Therefore,
                for any $n$,
                $s_n +dfrac1{4n+4}
                le S
                < s_n +dfrac1{4n}$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If
                  $s_n
                  = sum_{k=1}^nfrac{1}{n+k}
                  $,
                  then



                  $begin{array}\
                  s_{n+1}
                  &= sum_{k=1}^{n+1}dfrac{1}{n+1+k}\
                  &= sum_{k=2}^{n+2}dfrac{1}{n+k}\
                  &= sum_{k=2}^{n}dfrac{1}{n+k}+dfrac1{2n+1}+dfrac1{2n+2}\
                  &= sum_{k=1}^{n}dfrac{1}{n+k}-dfrac1{n+1}+dfrac1{2n+1}+dfrac1{2n+2}\
                  &= s_n+dfrac{(2n+1)+(2n+2)-2(2n+1)}{(2n+1)(2n+2)}\
                  &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                  end{array}
                  $



                  Therefore
                  $s_{n+1} > s_n$,
                  so $s_n$
                  is increasing.



                  Therefore,
                  if $n > m$,
                  $s_n > s_m$.



                  Note that
                  we can bound $s_n$ because



                  $begin{array}\
                  s_{n+1}
                  &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                  &< s_n+dfrac{1}{(2n)(2n+2)}\
                  &< s_n+frac14dfrac{1}{n(n+1)}\
                  &< s_n+frac14(dfrac1{n}-dfrac1{n+1})\
                  end{array}
                  $



                  so that
                  $s_{n+1}-s_n
                  < frac14(dfrac1{n}-dfrac1{n+1})
                  $.



                  Summing,
                  $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
                  < sum_{k=1}^{m-1}frac14(dfrac1{n}-dfrac1{n+1})
                  $
                  or
                  $s_{n+m}-s_n
                  lt frac14(frac1{n}-frac1{n+m})
                  lt frac1{4n}
                  $.



                  Therefore,
                  $lim_{n to infty} s_n$
                  exists,
                  and,
                  if
                  $S=lim_{n to infty} s_n$,
                  $S < s_n +dfrac1{4n}$.



                  We can similarly get a
                  lower bound on $s_n$
                  and, therefore,
                  a lower bound on $S$
                  in terms of $n$ and $s_n$.



                  (the following is done
                  with copy, paste, and edit.)



                  $begin{array}\
                  s_{n+1}
                  &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                  &> s_n+dfrac{1}{(2n+2)(2n+4)}\
                  &> s_n+frac14dfrac{1}{(n+2)(n+2)}\
                  &> s_n+frac14(dfrac1{n+1}-dfrac1{n+2})\
                  end{array}
                  $



                  so that
                  $s_{n+1}-s_n
                  gt frac14(dfrac1{n+1}-dfrac1{n+2})
                  $.



                  Summing,
                  $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
                  gt sum_{k=1}^{m-1}frac14(dfrac1{n+1}-dfrac1{n+2})
                  $
                  or
                  $s_{n+m}-s_n
                  gt frac14(frac1{n+1}-frac1{n+m+1})
                  $
                  so that
                  $S
                  =lim_{m to infty}s_{n+m}
                  ge s_n+lim_{m to infty}frac14(frac1{n+1}-frac1{n+m+1})
                  = s_n+frac1{4n+4}
                  $.



                  Therefore
                  $S ge s_n +dfrac1{4n+4}$.



                  Therefore,
                  for any $n$,
                  $s_n +dfrac1{4n+4}
                  le S
                  < s_n +dfrac1{4n}$.






                  share|cite|improve this answer









                  $endgroup$



                  If
                  $s_n
                  = sum_{k=1}^nfrac{1}{n+k}
                  $,
                  then



                  $begin{array}\
                  s_{n+1}
                  &= sum_{k=1}^{n+1}dfrac{1}{n+1+k}\
                  &= sum_{k=2}^{n+2}dfrac{1}{n+k}\
                  &= sum_{k=2}^{n}dfrac{1}{n+k}+dfrac1{2n+1}+dfrac1{2n+2}\
                  &= sum_{k=1}^{n}dfrac{1}{n+k}-dfrac1{n+1}+dfrac1{2n+1}+dfrac1{2n+2}\
                  &= s_n+dfrac{(2n+1)+(2n+2)-2(2n+1)}{(2n+1)(2n+2)}\
                  &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                  end{array}
                  $



                  Therefore
                  $s_{n+1} > s_n$,
                  so $s_n$
                  is increasing.



                  Therefore,
                  if $n > m$,
                  $s_n > s_m$.



                  Note that
                  we can bound $s_n$ because



                  $begin{array}\
                  s_{n+1}
                  &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                  &< s_n+dfrac{1}{(2n)(2n+2)}\
                  &< s_n+frac14dfrac{1}{n(n+1)}\
                  &< s_n+frac14(dfrac1{n}-dfrac1{n+1})\
                  end{array}
                  $



                  so that
                  $s_{n+1}-s_n
                  < frac14(dfrac1{n}-dfrac1{n+1})
                  $.



                  Summing,
                  $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
                  < sum_{k=1}^{m-1}frac14(dfrac1{n}-dfrac1{n+1})
                  $
                  or
                  $s_{n+m}-s_n
                  lt frac14(frac1{n}-frac1{n+m})
                  lt frac1{4n}
                  $.



                  Therefore,
                  $lim_{n to infty} s_n$
                  exists,
                  and,
                  if
                  $S=lim_{n to infty} s_n$,
                  $S < s_n +dfrac1{4n}$.



                  We can similarly get a
                  lower bound on $s_n$
                  and, therefore,
                  a lower bound on $S$
                  in terms of $n$ and $s_n$.



                  (the following is done
                  with copy, paste, and edit.)



                  $begin{array}\
                  s_{n+1}
                  &= s_n+dfrac{1}{(2n+1)(2n+2)}\
                  &> s_n+dfrac{1}{(2n+2)(2n+4)}\
                  &> s_n+frac14dfrac{1}{(n+2)(n+2)}\
                  &> s_n+frac14(dfrac1{n+1}-dfrac1{n+2})\
                  end{array}
                  $



                  so that
                  $s_{n+1}-s_n
                  gt frac14(dfrac1{n+1}-dfrac1{n+2})
                  $.



                  Summing,
                  $sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k})
                  gt sum_{k=1}^{m-1}frac14(dfrac1{n+1}-dfrac1{n+2})
                  $
                  or
                  $s_{n+m}-s_n
                  gt frac14(frac1{n+1}-frac1{n+m+1})
                  $
                  so that
                  $S
                  =lim_{m to infty}s_{n+m}
                  ge s_n+lim_{m to infty}frac14(frac1{n+1}-frac1{n+m+1})
                  = s_n+frac1{4n+4}
                  $.



                  Therefore
                  $S ge s_n +dfrac1{4n+4}$.



                  Therefore,
                  for any $n$,
                  $s_n +dfrac1{4n+4}
                  le S
                  < s_n +dfrac1{4n}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 26 '17 at 20:45









                  marty cohenmarty cohen

                  73.1k549128




                  73.1k549128






























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