Is $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$ equal to $\frac{1}{2n+2}$? On p. 146 of Principles of Mathematical Analysis, Rudin claims that $\int_{0}^{1} x(1-x^2)^ndx=\frac{1}{2n+2}$. But evaluating the integral I found it equal to $$\int_{0}^{1}x(\sum\limits_{k=0}^{n}{{n}\choose{k}}(-x^2)^k)dx=\int_{0}^{1}\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^{2k+1}dx=\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}\frac{x^{2k+2}}{2k+2}$$ evaluated at $x=1$, which equals $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$. So now I'm just wondering, why is this equal to $\frac{1}{2n+2}$?
 A: $$\begin{align*}
\sum_{k=0}^n\binom{n}k\frac{(-1)^k}{2k+2}&=\frac{1}2\sum_{k=0}^n\frac{1}{k+1}\binom{n}k(-1)^k\\
&=\frac{1}2\sum_{k=0}^n\frac{1}{n+1}\binom{n+1}{k+1}(-1)^k\\
&=\frac{1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^{k-1}\\
&=\frac{-1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}\\
&=\frac{-1}{2n+2}\left(\sum_{k=0}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}-\binom{n+1}0(-1)^01^{n+1}\right)\\
&=\frac{-1}{2n+2}\left((-1+1)^{n+1}-1\right)\\
&=\frac{1}{2n+2}
\end{align*}$$
by the binomial theorem.
A: HINT:
Note that we can write
$$\frac{1}{2k+2}\,\binom{n}{k}=\frac{1}{2n+2}\binom{n+1}{k+1}$$
Then, substitute this into the summation, shift the index from $k$ to $k'=k+1$ so the new summation limits are $k'=1$ to $k'=n+1$, and use $\sum\limits_{k=0}^{n+1} \binom{n}{k}(-1)^k=0$.
A: Why not use the substitution $1-x^{2}=t$ to change the integral to $$\frac{1}{2}\int_{0}^{1}t^{n}\,dt=\frac{1}{2n+2}$$ But you seem more interested in the link between your sum and value of integral in a direct manner as given in answers by Brian Scott and Dr. MV. 
A: An overkill. The Melzak's identity tell us that $$f\left(x+y\right)=y\dbinom{y+n}{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{f\left(x-k\right)}{y+k}
 $$ where $f$ is an algebraic polynomial up to dergee $n$, x,y\in\mathbb{R},\, y\neq-1,\dots-n
 . So taking $f\equiv1
 $ and $y=1$ we get $$1=\dbinom{1+n}{n}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{1+k}
 $$ hence $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{1+k}=\color{red}{\frac{1}{n+1}}\tag{1}
 $$ now you have only to multiply both side of $(1)$ by $1/2$.
