Alternating sum of binomial coefficients $\sum(-1)^k{n\choose k}\frac{1}{k+1}$ I would appreciate if somebody could help me with the following problem
Q:Calculate the sum:
$$ \sum_{k=1}^n (-1)^k {n\choose k}\frac{1}{k+1} $$

$$
\begin{align}
\sum_{k=1}^n(-1)^k\binom{n}{k}\frac1{k+1}
&=\sum_{k=1}^n(-1)^k\binom{n+1}{k+1}\frac1{n+1}\\
&=\frac1{n+1}\sum_{k=2}^{n+1}(-1)^{k-1}\binom{n+1}{k}\\
&=\frac1{n+1}\left(1-(n+1)+\sum_{k=0}^{n+1}(-1)^{k-1}\binom{n+1}{k}\right)\\
&=-\frac n{n+1}
\end{align}
$$
