Evaluating series with factorial denominator (sanity check).












9












$begingroup$


Is my approach to evaluating this series correct?



$$sum_{n=1}^infty frac{n}{(n+1)!}$$



Has partial sum equivalent to:



$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$



For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving



$$ S_m =1-frac{1}{(m+1)!}$$



Hence $ lim_{mrightarrowinfty} S_m = 1$



Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!






sequences-and-series







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$endgroup$












  • $begingroup$
    Looks fine to me.
    $endgroup$
    – Brian M. Scott
    Jan 5 '13 at 23:59










  • $begingroup$
    I see nothing wrong. I'm editing to improve formatting, however.
    $endgroup$
    – Alex Becker
    Jan 6 '13 at 0:03










  • $begingroup$
    It should be $S_m$ everywhere
    $endgroup$
    – leo
    Jan 6 '13 at 0:04
















9












$begingroup$


Is my approach to evaluating this series correct?



$$sum_{n=1}^infty frac{n}{(n+1)!}$$



Has partial sum equivalent to:



$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$



For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving



$$ S_m =1-frac{1}{(m+1)!}$$



Hence $ lim_{mrightarrowinfty} S_m = 1$



Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!






sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Looks fine to me.
    $endgroup$
    – Brian M. Scott
    Jan 5 '13 at 23:59










  • $begingroup$
    I see nothing wrong. I'm editing to improve formatting, however.
    $endgroup$
    – Alex Becker
    Jan 6 '13 at 0:03










  • $begingroup$
    It should be $S_m$ everywhere
    $endgroup$
    – leo
    Jan 6 '13 at 0:04














9












9








9





$begingroup$


Is my approach to evaluating this series correct?



$$sum_{n=1}^infty frac{n}{(n+1)!}$$



Has partial sum equivalent to:



$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$



For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving



$$ S_m =1-frac{1}{(m+1)!}$$



Hence $ lim_{mrightarrowinfty} S_m = 1$



Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!






sequences-and-series







share|cite|improve this question











$endgroup$




Is my approach to evaluating this series correct?



$$sum_{n=1}^infty frac{n}{(n+1)!}$$



Has partial sum equivalent to:



$$S_m = sum_{n=1}^m frac{n}{(n+1)!} = sum_{j=2}^{m+1} frac{j-1}{j!} = sum_{j=2}^{m+1} frac{1}{(j-1)!} - sum_{j=2}^{m+1} frac{1}{j!} $$



For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving



$$ S_m =1-frac{1}{(m+1)!}$$



Hence $ lim_{mrightarrowinfty} S_m = 1$



Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 '13 at 0:19









Calvin Lin

36.3k349114




36.3k349114










asked Jan 5 '13 at 23:54









conjecturesconjectures

534513




534513












  • $begingroup$
    Looks fine to me.
    $endgroup$
    – Brian M. Scott
    Jan 5 '13 at 23:59










  • $begingroup$
    I see nothing wrong. I'm editing to improve formatting, however.
    $endgroup$
    – Alex Becker
    Jan 6 '13 at 0:03










  • $begingroup$
    It should be $S_m$ everywhere
    $endgroup$
    – leo
    Jan 6 '13 at 0:04


















  • $begingroup$
    Looks fine to me.
    $endgroup$
    – Brian M. Scott
    Jan 5 '13 at 23:59










  • $begingroup$
    I see nothing wrong. I'm editing to improve formatting, however.
    $endgroup$
    – Alex Becker
    Jan 6 '13 at 0:03










  • $begingroup$
    It should be $S_m$ everywhere
    $endgroup$
    – leo
    Jan 6 '13 at 0:04
















$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59




$begingroup$
Looks fine to me.
$endgroup$
– Brian M. Scott
Jan 5 '13 at 23:59












$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03




$begingroup$
I see nothing wrong. I'm editing to improve formatting, however.
$endgroup$
– Alex Becker
Jan 6 '13 at 0:03












$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04




$begingroup$
It should be $S_m$ everywhere
$endgroup$
– leo
Jan 6 '13 at 0:04










2 Answers
2






active

oldest

votes


















4












$begingroup$

Your solution is not only correct but also a very nice one.



Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step



$$
begin{align}
S_{m+1}&=S_m+frac{m+1}{(m+2)!}
\
&=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
\
&=1+frac{m+1-(m+2)}{(m+2)!}
\
&=1-frac1{(m+2)!};.
end{align}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Cheers Joriki - hadn't actually thought of trying it by induction.
    $endgroup$
    – conjectures
    Jan 6 '13 at 10:14



















-1












$begingroup$

$$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
$$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Your solution is not only correct but also a very nice one.



    Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step



    $$
    begin{align}
    S_{m+1}&=S_m+frac{m+1}{(m+2)!}
    \
    &=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
    \
    &=1+frac{m+1-(m+2)}{(m+2)!}
    \
    &=1-frac1{(m+2)!};.
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Cheers Joriki - hadn't actually thought of trying it by induction.
      $endgroup$
      – conjectures
      Jan 6 '13 at 10:14
















    4












    $begingroup$

    Your solution is not only correct but also a very nice one.



    Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step



    $$
    begin{align}
    S_{m+1}&=S_m+frac{m+1}{(m+2)!}
    \
    &=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
    \
    &=1+frac{m+1-(m+2)}{(m+2)!}
    \
    &=1-frac1{(m+2)!};.
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Cheers Joriki - hadn't actually thought of trying it by induction.
      $endgroup$
      – conjectures
      Jan 6 '13 at 10:14














    4












    4








    4





    $begingroup$

    Your solution is not only correct but also a very nice one.



    Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step



    $$
    begin{align}
    S_{m+1}&=S_m+frac{m+1}{(m+2)!}
    \
    &=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
    \
    &=1+frac{m+1-(m+2)}{(m+2)!}
    \
    &=1-frac1{(m+2)!};.
    end{align}
    $$






    share|cite|improve this answer









    $endgroup$



    Your solution is not only correct but also a very nice one.



    Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step



    $$
    begin{align}
    S_{m+1}&=S_m+frac{m+1}{(m+2)!}
    \
    &=1-frac1{(m+1)!}+frac{m+1}{(m+2)!}
    \
    &=1+frac{m+1-(m+2)}{(m+2)!}
    \
    &=1-frac1{(m+2)!};.
    end{align}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 6 '13 at 8:02









    jorikijoriki

    170k10183343




    170k10183343












    • $begingroup$
      Cheers Joriki - hadn't actually thought of trying it by induction.
      $endgroup$
      – conjectures
      Jan 6 '13 at 10:14


















    • $begingroup$
      Cheers Joriki - hadn't actually thought of trying it by induction.
      $endgroup$
      – conjectures
      Jan 6 '13 at 10:14
















    $begingroup$
    Cheers Joriki - hadn't actually thought of trying it by induction.
    $endgroup$
    – conjectures
    Jan 6 '13 at 10:14




    $begingroup$
    Cheers Joriki - hadn't actually thought of trying it by induction.
    $endgroup$
    – conjectures
    Jan 6 '13 at 10:14











    -1












    $begingroup$

    $$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
    $$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$






    share|cite|improve this answer









    $endgroup$


















      -1












      $begingroup$

      $$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
      $$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$






      share|cite|improve this answer









      $endgroup$
















        -1












        -1








        -1





        $begingroup$

        $$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
        $$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$






        share|cite|improve this answer









        $endgroup$



        $$sum_{n=1}^infty frac{n}{(n+1)!}$$= $$sum_{n=1}^infty frac{n+1-1}{(n+1)!}$$=
        $$sum_{n=1}^infty frac{1}{n!}-frac{1}{(n+1)!}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 15:15









        Kostas GiatzoKostas Giatzo

        12




        12






























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