Prove $\sum_{k=0}^{n}\frac{\binom{n}{k}(-1)^k}{k+1}$ = $\frac{1}{n+1}$ Any tips on where to start? I tried induction, using the inductive property of Binomial coefficients and the Mean Value Theorem for divided differences however I haven't made any progress.
 A: We have
$$f(x)=\sum_{k=0}^n {n\choose k}(-1)^k x^k=(1-x)^n$$
and the desired sum is 
$$\int_0^1f(x)dx=\int_0^1(1-x)^ndx=-\frac1{n+1}(1-x)^{n+1}\Bigg|_0^1=\frac1{n+1}$$
A: *

*Use $\displaystyle\int_{0}^{1} x^{k} \ dx = \frac{1}{k+1}$

A: Here's a non-calculus argument.
\begin{align*}
\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k+1} &= \sum_{k=0}^n (-1)^k\frac{n!}{k!(n-k)!}\frac{1}{k+1}\\
&= \sum_{k=0}^n (-1)^k \frac{1}{n+1}\frac{(n+1)!}{(k+1)!(n-k)!}\\
&= \frac{1}{n+1}\sum_{k=0}^n (-1)^k \binom{n+1}{k+1}\\
&= \frac{1}{n+1}\sum_{k=0}^n (-1)^k \bigg( \binom{n}{k} + \binom{n}{k+1}\bigg)\\
&= \frac{1}{n+1}
\end{align*}
since the sum telescopes and all the terms but $\binom{n}{0} = 1$ cancel.
To try to demystify this --- The idea that inspired this line of thinking was observing that the common denominator of these fractions is $(n+1)!$ and wondering if it would be possible to put the sum in the numerator of a fraction with denominator $n+1$. Then I noticed that $(k+1)k! = (k+1)!$ and it clicked.

Going from line $2$ to line $3$, 
$$ \binom{n+1}{k+1} = \frac{ (n+1)!}{ (k+1)![ (n+1) - (k+1) ]!} = \frac{(n+1)!}{(k+1)!(n-k)!}.$$
