Prove that $\sum_{m=1}^{n} (-1)^{m+1} {n \choose m} \frac{1}{m+1} = \frac{n}{n+1}$

Question

I'm trying to prove that:

$$\sum_{m=1}^{n} (-1)^{m+1} {n \choose m} \frac{1}{m+1} = \frac{n}{n+1}$$

I've tried to prove this by induction and directly, without luck. Any help would be appreciated.

Answer 1

HINT: Use the identity

$$\binom{n}m\frac1{m+1}=\binom{n+1}{m+1}\frac1{n+1}\,.$$

I’ve taken it a step further in the spoiler block below.

$$\begin{align*}\sum_{m=1}^n(-1)^{m+1}\binom{n}m\frac1{m+1}&=\frac1{n+1}\sum_{m=1}^n(-1)^{m+1}\binom{n+1}{m+1}\\&=\frac1{n+1}\sum_{k=2}^{n+1}(-1)^k\binom{n+1}k\end{align*}$$