<s>[INST] Closed form for a sum involving binomial coefficient $\sum_{j=0}^n \binom{n}{j} \frac1{j+1} = \frac{2^{n+1}-1}{n+1}$ 
Possible Duplicate:
How can I compute $\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} $? 

How to derive the following equality?
$$\sum_{j=0}^n \binom{n}{j} \frac1{j+1} = \frac{2^{n+1}-1}{n+1}$$
 [/INST] Using the identity
$$
\binom{n}{j}\frac{n+1}{j+1}=\binom{n+1}{j+1}\tag{1}
$$
and summing
$$
\begin{align}
\sum_{j=0}^n\binom{n}{j}\frac{n+1}{j+1}
&=\sum_{j=0}^n\binom{n+1}{j+1}\\
&=2^{n+1}-1\tag{2}
\end{align}
$$
Dividing by $n+1$ yields
$$
\sum_{j=0}^n\binom{n}{j}\frac{1}{j+1}=\frac{2^{n+1}-1}{n+1}\tag{3}
$$
</s>