Help ending a proof using binomial theorem: $\sum_{k=0}^n {n \choose k}\cdot\frac{(-1)^k}{k+1} = \frac{1}{n+1}$ I need to prove the following: $$\sum_{k=0}^n {n \choose k}\cdot\frac{(-1)^k}{k+1} = \frac{1}{n+1}$$
Started from the left, I got this far:  $$\frac{1}{n+1}+\sum_{k=1}^n {n+1 \choose k+1}\cdot(-1)^k$$
Am I right so far? Any ideas how to move on? thanks.
Edit: I guess my last move had a mistake, then perhaps this is ok?
$$\frac{1}{n+1}\cdot\sum_{k=0}^n {n+1 \choose k+1}\cdot(-1)^k$$
How do you suggest I move on?
 A: Integrate both sides of  $$\sum_{k=0}^n{n\choose k}x^k=(x+1)^n$$
to obtain that $$C+\sum_{k=0}^n\frac{1}{k+1}{n\choose k}x^{k+1}=\frac{1}{n+1}(x+1)^{n+1}$$
Where $C$ is a constant of integration which equals $\frac{1}{n+1}$ upon substituting $x=0$. For your identity, substitute $x=-1$.
A: Note that we have
$$
\binom{n}{k}\frac{(-1)^k}{k+1} = \frac{n!}{k!(n-k)!}\frac{(-1)^k}{k+1} = (-1)^k\frac{n!}{(k+1)!(n-k)!} =\\
 \frac{(-1)^k}{n+1}\frac{(n+1)!}{(k+1)!(n-k)!} = \binom{n+1}{k+1}\frac{(-1)^k}{n+1}\\
= \frac{-1}{n+1}\cdot (-1)^{k+1}\binom{n+1}{k+1}
$$
Now, while keeping the indices straight, compare the sum of these terms to what the binomial theorem says about
$$
\frac{(1-1)^{n+1}}{n+1}
$$

Edit: More details.
The binomial theorem says that
$$
0 = \frac{(1-1)^{n+1}}{n+1} = \sum_{m = 0}^{n+1}\frac{1}{n+1}\binom{n+1}{m}\tag1(-1)^{m}
$$
while after the transformation at the top of my answer, your sum looks like
$$
\sum_{k = 0}^n\frac{-1}{n+1}\binom{n+1}{k+1}(-1)^{k+1}
$$
Doing the index transformation $m = k+1$, this last one becomes
$$
\sum_{m = 1}^{n+1}\frac{-1}{n+1}\binom{n+1}{m}(-1)^{m}\tag2
$$
Now can you compare the sums $(1)$ and $(2)$?
