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?

Apparently I'm a little late to the party, but my answer has a punchline!
We have
$$
\frac{1}{z} \int_0^z \sum_{k=0}^{n} \binom{n}{k} s^k\,ds = \sum_{k=0}^{n} \binom{n}{k} \frac{z^k}{k+1},
$$
so that
$$
- \int_0^z \frac{1}{t} \int_0^t \sum_{k=0}^{n} \binom{n}{k} s^k\,ds\,dt = - \sum_{k=0}^{n} \binom{n}{k} \frac{z^{k+1}}{(k+1)^2}.
$$
Setting $z = -1$ gives an expression for your sum,
$$
\sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^k}{(k+1)^2} = \int_{-1}^{0} \frac{1}{t} \int_0^t \sum_{k=0}^{n} \binom{n}{k} s^k\,ds\,dt.
$$
Now, $\sum_{k=0}^{n} \binom{n}{k} s^k = (1+s)^n$, so
$$
\begin{align*}
\sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^k}{(k+1)^2} &= \int_{-1}^{0} \frac{1}{t} \int_0^t (1+s)^n \,ds\,dt \\
&= \frac{1}{n+1}\int_{-1}^{0} \frac{1}{t} \left[(1+t)^{n+1} - 1\right]\,dt \\
&= \frac{1}{n+1}\int_{0}^{1} \frac{u^{n+1}-1}{u-1}\,du \\
&= \frac{1}{n+1}\int_0^1 \sum_{k=0}^{n} u^k \,du \\
&= \frac{1}{n+1}\sum_{k=1}^{n+1} \frac{1}{k} \\
&= \frac{H_{n+1}}{n+1},
\end{align*}
$$
where $H_n$ is the $n^{\text{th}}$ harmonic number.
