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 $m+1&gt;j&gt;2$ 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!
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

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