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_{m\rightarrow\infty} S_m = 1$

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

  • $\begingroup$ 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

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

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

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

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

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