How would I prove this strange combinatorial identity: $\sum_{i=0}^m \binom mi (-1)^i \frac{1}{i+1} = \frac{1}{m+1}$?

Question

I came across a combinatorial identity today but have no idea how to handle the $\frac{1}{i+1}$ term:

$$\sum_{i=0}^m {m \choose i} (-1)^i \frac{1}{i+1} = \frac{1}{m+1}$$

How would I prove it?

Answer 1

$$\begin{align}\sum_{i=0}^{m}\color{red}{\binom{m}{i}}(-1)^i\color{red}{\frac{1}{i+1}}&=-\sum_{i=0}^{m}\color{red}{\frac{1}{m+1}\binom{m+1}{i+1}}(-1)^{i+1}\\&=-\frac{1}{m+1}\left(-1+\sum_{j=0}^{m+1}\binom{m+1}{j}(-1)^j\right)\\&=-\frac{1}{m+1}\left(-1+(1-1)^{m+1}\right)\\&=\frac{1}{m+1}\end{align}$$

Answer 2

Recall that $$\sum_{i=0}^m \dbinom{m}i x^i = (1+x)^m$$ Integrate the above from $-1$ to $0$. This gives us $$\sum_{i=0}^m \dbinom{m}i \left.\dfrac{x^{i+1}}{i+1}\right \vert_{-1}^0 = \dfrac{(1+x)^{m+1}}{m+1}$$ which is what you want.

Answer 3

\begin{eqnarray} \sum_{i=0}^m{m\choose i}\frac{(-1)^i}{i+1}&=&\sum_{i=0}^m(-1)^i\frac{m!}{(m-i)!(i+1)!}=\sum_{i=0}^m(-1)^i\frac{m+1\choose i+1}{m+1}=\frac{1}{m+1}\sum_{i=0}^m(-1)^i{m+1\choose i+1}\\ &=&\frac{-1}{m+1}\sum_{i=1}^{m+1}(-1)^i{m+1\choose i}=\frac{1}{m+1}-\frac{1}{m+1}\sum_{i=0}^{m+1}(-1)^i{m+1\choose i}\\ &=&\frac{1}{m+1}-\frac{1}{m+1}(1-1)^{m+1}=\frac{1}{m+1}. \end{eqnarray}