Finding a closed form for $\sum^{n}_{k=1} \frac{k}{(k+1)!} $ I'm finding a closed form to  $\sum^{n}_{k=1} \frac{k}{(k+1)!}  (n \leq 1) $ (in a environment of induction and recurrence) 
I've been trying to solve it without success, can anybody help me (?)
The last thing I've tried to do is write the expression as a Non-Homogeneous Linear Recurrence: 
$ f(1) = \frac{1}{2} $ 
but, actually I have no idea of how to attach it. 
 $ f(n+1) = f(n) + \frac{1}{(n+2)n!} $ 
 A: It is a telescopic series:
$$\sum_{k\geq 1}\frac{k}{(k+1)!}=\sum_{k\geq 1}\frac{(k+1)}{(k+1)!}-\sum_{k\geq 1}\frac{1}{(k+1)!}=\sum_{k\geq 1}\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)=\color{red}{1}$$
since:
$$ \sum_{k=1}^{N}\frac{k}{(k+1)!} = 1-\frac{1}{(N+1)!}.$$
A: prove by induction that $$\sum_{k=1}^n\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$$
A: Jack's answer is great, but if you are having difficult seeing the "since" portion of his answer, then consider the main portion of the inductive argument for establishing the identity that Jack gives:
\begin{align}
\sum_{k=1}^{n+1}\frac{k}{(k+1)!} &= \frac{n+1}{(n+2)!}+\color{green}{\sum_{k=1}^n\frac{k}{(k+1)!}}\tag{by defn. of $\Sigma$}\\[1em]
&= \frac{n+1}{(n+2)!}+\color{green}{\left[1-\frac{1}{(n+1)!}\right]}\tag{by ind. hyp.}\\[1em]
&= 1+\frac{n+1}{(n+2)!}-\frac{(n+2)}{(n+2)!}\tag{manipulate}\\[1em]
&= 1-\frac{1}{(n+2)!}\tag{closed form}
\end{align}
Note that the portion highlighted in green is where you use the inductive hypothesis. 
A: This is a supplement to @JackDaurizio's since part which was asked by OP in a comment.

It's all about telescoping in Jack's answer.
\begin{align*}
\sum_{k=1}^N\frac{k}{(k+1)!}&=\sum_{k=1}^N\frac{(k+1)-1}{(k+1)!}\\
&=\sum_{k=1}^N\left(\frac{1}{k!}-\frac{1}{(k+1)!}\right)\\
&=1-\frac{1}{(N+1)!}\\
\end{align*}
  It follows
  \begin{align*}
\sum_{k=1}^\infty\frac{k}{(k+1)!}&=\lim_{N\rightarrow\infty}\sum_{k=1}^N\frac{k}{(k+1)!}\\
&=\lim_{N\rightarrow\infty}\left(1-\frac{1}{(N+1)!}\right)\\
&=1
\end{align*}

