TITLE: Alternating sum of binomial coefficients multiplied by (1/k+1)
QUESTION [7 upvotes]: I'm trying to prove that
$$\sum_{k=0}^n {n \choose k} (-1)^k \frac{1}{k+1} = \frac{1}{n+1}$$
So far I've tried induction (which doesn't really work at all), using well known facts such as
$$\sum_{k=0}^n {n \choose k} (-1)^k = 0$$
and trying to apply identities like
$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$
Would anyone be able to point me towards the right method? Should I be looking to apply an identity or is there a method I'm missing?

REPLY [3 votes]: From $(k+1)\binom{n+1}{k+1} = (n+1)\binom{n}{k}$ we get
\begin{align*} 
\sum_{k=0}^n \binom{n}{k} (-1)^k \frac{1}{k+1} &= \frac{1}{n+1} \sum_{k=0}^n \binom{n+1}{k+1}(-1)^k \\
&= \frac{1}{n+1} \Bigl( 1 + \sum_{r=0}^{n+1} \binom{n+1}{r} (-1)^{r-1} \Bigr)  \\
&= \frac{1}{n+1}
\end{align*}
as required.