Proving $ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} (2^{n+1} - 1) $

Question

I'm stuck at proving the following.

$$ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = \frac{1}{n+1} (2^{n+1} - 1) $$

This is what I have so far.

$ \sum_{r=0}^{n} \frac{1}{r+1} \binom{n}{r} = (1) \binom{n}{0} + \frac{1}{2} \binom{n}{0} + \frac{1}{3} \binom{n}{2} + ... + \frac{1}{n+1} \binom{n}{n}$

Answer 1

Sice we have $$\frac{1}{r+1}\binom{n}{r}=\frac{1}{n+1}\binom{n+1}{r+1},$$ we have $$\begin{align}\sum_{r=0}^{n}\frac{1}{r+1}\binom{n}{r}&=\frac{1}{n+1}\sum_{r=0}^{n}\binom{n+1}{r+1}\\&=\frac{1}{n+1}\left(\sum_{r=0}^{n+1}\binom{n+1}{r}-1\right)\\&=\frac{1}{n+1}\left(2^{n+1}-1\right).\end{align}$$

Here, note that $$\sum_{r=0}^{n+1}\binom{n+1}{r}=\sum_{r=0}^{n+1}\binom{n+1}{r}\cdot 1^{n+1-r}\cdot 1^r=(1+1)^{n+1}=2^{n+1}.$$

Answer 2

A start: Let $f(t)=(1+t)^n=\sum_0^n \binom{n}{r}t^r$. Calculate $\int_0^1 f(t)\,dt$ in two different ways.