# Evaluating series with factorial denominator (sanity check).

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!

• Looks fine to me. Jan 5, 2013 at 23:59
• I see nothing wrong. I'm editing to improve formatting, however. Jan 6, 2013 at 0:03
• It should be $S_m$ everywhere
– leo
Jan 6, 2013 at 0:04

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