<s>[INST] Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$ How can I calculate the following sum involving binomial terms:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Where the value of n can get very big (thus calculating the binomial coefficients is not feasible).
Is there a closed form for this summation?
 [/INST] I'm even later to the party, but that's only because "absorption identity" kept yelling in my ear. :)
One application of the absorption identity gets one of the factors of $k+1$ out of the denominator:
$$\begin{align}
\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(k+1)^2} &= \frac{1}{n+1} \sum_{k=0}^n \binom{n+1}{k+1} \frac{(-1)^k}{k+1} \\
&= \frac{1}{n+1} \sum_{k=1}^{n+1} \binom{n+1}{k} \frac{(-1)^{k+1}}{k} \end{align}.$$
It would be nice to use the absorption identity again, but we need a $k+1$ in the denominator of the summand rather than a $k$.  By using the basic binomial coefficient recursion formula, we can make that happen.
Let $\displaystyle f(n) = \sum_{k=1}^{n} \binom{n}{k} \frac{(-1)^{k+1}}{k}.$  Then looking at the difference of $f(n+1)$ and $f(n)$ gives us
$$\begin{align}
f(n+1) - f(n) &= \sum_{k=1}^{n+1} \binom{n+1}{k} \frac{(-1)^{k+1}}{k} - \sum_{k=1}^n \binom{n}{k} \frac{(-1)^{k+1}}{k} \\
&= \sum_{k=1}^{n+1} \binom{n}{k-1} \frac{(-1)^{k+1}}{k} \\
&= \sum_{k=0}^n \binom{n}{k} \frac{(-1)^{k}}{k+1} \\
&= \frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k+1} (-1)^{k} \:\:\:\: \text{ (absorption identity!)} \\
&= \frac{1}{n+1} \left(1 + \sum_{k=0}^{n+1} \binom{n+1}{k} (-1)^{k+1} \right)\\
&= \frac{1}{n+1},
\end{align}$$
where in the last step we used the fact that the alternating sum of the binomial coefficients is $0$.
Thus $$f(n+1) = \sum_{k=0}^n (f(k+1) - f(k)) = \sum_{k=0}^n \frac{1}{k+1} = H_{n+1}.$$
Therefore,
$$\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(k+1)^2} = \frac{H_{n+1}}{n+1}.$$
</s>