Compute the value of $$\sum_{k=0}^{n}(-1)^k\frac{1}{k+1}\binom{n}{k}=\binom{n}{0}-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2} -\cdots+ (-1)^n\frac{1}{n+1}\binom{n}{n}$$
Applying the basic identity $\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}$ gives $$\begin{align*} &\binom{n}{0}-\frac{1}{2}\binom{n}{1}+ \frac{1}{3}\binom{n}{2} -\cdots + (-1)^n\frac{1}{n+1}\binom{n}{n}\\ \\ =\ &\frac{1}{n+1}\binom{n+1}{1} - \frac{1}{n+1}\binom{n+1}{2} +\cdots +(-1)^n\frac{1}{n+1}\binom{n+1}{n+1} \\ \\=\ & \frac{1}{n+1}\left(-\binom{n+1}{0} + \binom{n+1}{1} - \cdots +(-1)^n\binom{n+1}{n+1}\right) +\frac{1}{n+1}\binom{n+1}{0} \\ \\=\ & 0 + \frac{1}{n+1} \\ \\=\ &\boxed{\frac{1}{n+1}} \end{align*}$$
The last step utilizes the result of # 3158.