Is $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$ equal to $\frac{1}{2n+2}$?

$$\begin{align*} \sum_{k=0}^n\binom{n}k\frac{(-1)^k}{2k+2}&=\frac{1}2\sum_{k=0}^n\frac{1}{k+1}\binom{n}k(-1)^k\\ &=\frac{1}2\sum_{k=0}^n\frac{1}{n+1}\binom{n+1}{k+1}(-1)^k\\ &=\frac{1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^{k-1}\\ &=\frac{-1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}\\ &=\frac{-1}{2n+2}\left(\sum_{k=0}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}-\binom{n+1}0(-1)^01^{n+1}\right)\\ &=\frac{-1}{2n+2}\left((-1+1)^{n+1}-1\right)\\ &=\frac{1}{2n+2} \end{align*}$$

by the binomial theorem.

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 $\sum\limits_{k=0}^{n+1} \binom{n}{k}(-1)^k=0$.

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.