Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$

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.

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, $$\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(k+1)^2} = \frac{H_{n+1}}{n+1}.$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{k = 0}^{n}{n \choose k}{\pars{-1}^{k} \over \pars{k + 1}^{2}}} \\[5mm] = &\ -\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k}\int_{0}^{1}\ln\pars{x}\,x^{k}\,\dd x \\[5mm] = &\ -\int_{0}^{1}\ln\pars{x}\sum_{k = 0}^{n}{n \choose k}\pars{-x}^{k}\,\dd x \\[5mm] & = -\int_{0}^{1}\ln\pars{x}\pars{1 - x}^{n}\,\dd x \\[5mm] = &\ \left.-\,\partiald{}{\mu}\int_{0}^{1}x^{\mu}\pars{1 - x}^{n} \,\dd x\, \right\vert_{\ \mu\ =\ 0} \\[5mm] & = \left.-\,\partiald{}{\mu} {\Gamma\pars{\mu + 1}\Gamma\pars{n + 1} \over \Gamma\pars{\mu + n + 2}} \,\right\vert_{\ \mu\ =\ 0}\ =\ \bbx{\ds{H_{n + 1} \over n + 1}} \\ & \end{align}

\begin{align} &\bbox[5px,#ffd]{% -\,{\Gamma\pars{\mu + 1}\Gamma\pars{n + 1} \over \Gamma\pars{\mu + n + 2}}} \\[5mm] = &\ -\,{n! \over \pars{n + 1}!}\ \times \\[2mm] &\ \braces{1 + \bracks{\pars{1 - \gamma}-\pars{H_{n + 1} - \gamma}}\mu + \,\mrm{O}\pars{\mu^{2}}} \\[5mm] = &\ -\,{1 \over n + 1} + \color{red}{H_{n} \over n + 1}\,\mu + \,\mrm{O}\pars{\mu^{2}}\ \mbox{with}\ H_{0} = 0. \end{align}

$(1-x)^n=\sum_{0\le k\le n}\binom nk(-1)^kx^k$

Integrating wrt $x,$

$$-\frac{(1-x)^{n+1}}{n+1}+C=\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k+1}}{k+1}$$ where $C$ is the indefinite constant.

Putting $x=0,C-\frac1{n+1}=0\implies C=\frac1{n+1}$

So, $$\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k+1}}{k+1}=\frac1{n+1}-\frac{(1-x)^{n+1}}{n+1}$$

So, $$\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k}}{k+1}=\frac1{x(n+1)}-\frac{(1-x)^{n+1}}{(n+1)x}$$

Again integrating wrt $x,$

$$\sum_{0\le k\le n}\binom nk(-1)^k\frac{x^{k+1}}{(k+1)^2}=\frac {\log x}{n+1}-\int\frac{(1-x)^{n+1}}{(n+1)x}dx+D$$ where $D$ is the indefinite constant.