TITLE: Evaluate: $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics.
QUESTION [5 upvotes]: 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

REPLY [2 votes]: Solution
Notice that$$\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}.$$
Hence, $$\sum_{k=1}^n \frac{k}{(k+1)!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)=1-\frac{1}{(n+1)!}.$$