Evaluate: $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics. Evaluate    $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$.  This is from a combinatorics textbook so I'd like a combinatorial proof.  I find doing this kind of problem difficult especially when you have to sum - I don't know how to construct a sensible analogy using the addition principle.  
Similar question that appears just before this question in the text: Combinatorics problem involving series summation

Using generating functions, which are widely used in combinatorics:
$$a_n=\sum\limits_{k=1}^{n}\frac{k}{(k+1)!}$$
which is the same as
$$a_n=a_{n-1}+\frac{n}{(n+1)!}$$
with generating function
$$f(x)=\sum\limits_{n}\color{red}{a_n}x^n
=a_0+\sum\limits_{n=1}\left(a_{n-1}+\frac{n}{(n+1)!}\right)x^n=\\
a_0+x\sum\limits_{n=1}a_{n-1}x^{n-1}+\sum\limits_{n=1}\frac{n}{(n+1)!}x^n=\\
a_0+xf(x)+\sum\limits_{n=1}\frac{n+1}{(n+1)!}x^n-\sum\limits_{n=1}\frac{1}{(n+1)!}x^n=\\
a_0+xf(x)+\left(\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}\right)'-\frac{1}{x}\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}=\\
a_0+xf(x)+\left(e^x-1-x\right)'-\frac{1}{x}\left(e^x-1-x\right)=\\
a_0+xf(x)+e^x-\frac{1}{x}\left(e^x-1\right)$$
or
$$f(x)=\frac{a_0}{1-x}+\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}$$
since $a_0=0$
$$f(x)=\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}=\frac{1}{(1-x)x}-\frac{e^x}{x}=\\
\frac{1}{x}\left(\sum\limits_{n=0}x^n - \sum\limits_{n=0}\frac{x^n}{n!}\right)=\sum\limits_{n=1}\color{red}{\left(1-\frac{1}{(n+1)!}\right)}x^{n}$$
as a result
$$a_n=1-\frac{1}{(n+1)!}, n\geq1$$
