Showing $\sum_0^n {n\choose k}(-1)^k \frac{1}{2k + 2} = 1/(2n + 2)$ I'm wondering if there is an natural way to see that
$\sum_0^n {n\choose k}(-1)^k \frac{1}{2k + 2} = 1/(2n + 2)$  for $n\gt 0$.
Both sides of the equation are equal to a simple integral Integral of $x(1-x^2)^n$ 
By "natural" I mean that you would be able to deduce the right hand side without knowing it previously. For example, an induction argument is slightly unnatural since we would have to guess the closed form beforehand.
 A: $$\binom nk\frac1{2(k+1)}=\frac12\cdot\frac{n!}{(n-k)!\cdot k!\cdot(k+1)}$$
$$=\frac1{2(n+1)}\cdot\frac{(n+1)!}{[n+1-(k+1)]!\cdot(k+1)!}=\frac1{2n+2}\binom{n+1}{k+1}$$
So, $\displaystyle\sum_{k=0}^n\binom nk(-1)^k\frac1{2k+2}=\frac1{2n+2}\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}$
If $\displaystyle S=\sum_{k=0}^n(-1)^k\binom{n+1}{k+1}=-\sum_{k=0}^n(-1)^{k+1}\binom{n+1}{k+1}$
Setting $k+1=r,$
$$S=-\sum_{r=1}^{n+1}(-1)^r\binom{n+1}r\iff-S=\sum_{r=1}^{n+1}(-1)^r\binom{n+1}r$$
$$-S=\sum_{r=0}^{n+1}(-1)^r\binom{n+1}r-\binom{n+1}0(-1)^0=(1-1)^{n+1}-1$$
A: $$(1+x)^n=\sum_{r=0}^n\binom nrx^r$$
Integrate both sides   to get $$\frac{(1+x)^{n+1}}{n+1} =\sum_{r=0}^n\binom nr\frac{x^{r+1}}{r+1}+K $$ where $K$ is an arbitrary constant 
Set $x=0$ in the above identity to determine $K$
Finally set $x=-1$ 
A: You want to prove
$$\sum_{k=0}^n {n\choose k} \frac{1}{k + 1} (-1)^k= \frac{1}{n + 1}$$
i.e.
$$\sum_{k=0}^n {n\choose k}\frac{n+1}{k + 1} (-1)^k = 1$$
which is 
$$\sum_{k=0}^n {n+1\choose k+1}(-1)^k  = 1$$
which is 
$$\frac{1}{X}\sum_{k=0}^{n+1} {n+1\choose k+1}X^{k+1}\left.\right|_{X=\,-1}  = 1$$
As 
$\ \ (X+Y)^n=\sum_{k=0}^n\binom nk X^{n-k}Y^k\ \ $ 
we find the left hand side is
$$\frac{1}{X}\left((X+1)^{n+1}-1\right)\left.\right|_{X=\,-1}$$
and that's $1$ indeed.
