How can I compute $\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} $? This sum is difficult.  How can I compute it, without using calculus? 

$$\sum_{k = 1}^n \frac1{k + 1}\binom{n}{k}$$

If someone can explain some technique to do it, I'd appreciate it.
Or advice using a telescopic sum, I think with a telescopic could go, but do not know how to assemble it.

\begin{align*}\sum_{k=1}^n \binom{n}{k} \frac{1}{k+1} 
= \frac{1}{n+1} \sum_{k=1}^n \binom{n+1}{k+1} 
= \frac{2^{n+1} - 1 - (n+1)}{n+1} = \frac{2^{n+1} - n-2}{n+1}.
\end{align*}
The first step follows from the identity $\binom{n}{k} \frac{n+1}{k+1} = \frac{n!}{k! (n-k)!} \frac{n+1}{k+1} = \frac{(n+1)!}{(k+1)! (n-k)!} = \binom{n+1}{k+1}$.  The second step uses the fact that $\sum_{k=0}^{n+1} \binom{n+1}{k} = 2^{n+1}$, while noting that $\binom{n+1}{0}$ and $\binom{n+1}{1}$ are not included in the sum.
