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?
$$\sum_{j=0}^n \binom{n}{j} \frac1{j+1} = \frac{2^{n+1}-1}{n+1}$$
 A: We have $$(1+x)^n = \sum_{j=0}^n \binom{n}{j}x^j.$$
Integrating both sides from $0$ to $x$, we see that 
$$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{j=0}^n \binom{n}{j}\frac{x^{j+1}}{j+1}.$$
Putting $x=1$ yields the desired expression.
A: Here's a probabilistic approach. We want to show that
$$
\frac{1}{2^{n+1}-1} \sum_{j=0}^n \binom{n}{j} \frac{1}{j+1} = \frac{1}{n+1}. \tag{$\ast$}
$$
Consider an experiment where we pick a nonempty subset ("committee") $S \subseteq \{ 0, 1, 2, \ldots, n \}$ uniformly at random and pick an $x$ uniformly at random from $S$ (the "head" of the committee). Then both sides count the probability that the head is $0$. By symmetry the head is a uniformly random person, so it is $n+1$ with probability $\frac{1}{n+1}$. Therefore, it only remains to justify the left hand side of $(\ast)$.
The probability of the event $E_j$ that $S$ contains $0$ and it also $j$ other people from $\{ 1, 2, \ldots, n\}$ is 
$$
\frac{1}{2^{n+1}-1} \binom{n}{j}.
$$
Conditioned on $E_j$, the probability that the head is $0$ is equal to $\frac{1}{j+1}$. [Finally, conditioned on the event that $0 \notin S$, the probability that the head is $0$ is zero.] Thus, by the law of total probability, 
$$
\begin{align*}
\Pr[\text{Head is } 0] &= \sum_{j=0}^n \Pr[E_j] \cdot \Pr[\text{Head is } 0 \mid E_j]
\\  &= \sum_{j=0}^n \frac{1}{2^{n+1}-1} \binom{n}{j} \cdot \frac{1}{j+1}
\end{align*}
$$
A: 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
$$
\sum_{j=0}^n\binom{n}{j}\frac{1}{j+1}=\frac{2^{n+1}-1}{n+1}\tag{3}
$$
