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?
 A: 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,
$$\sum_{k=0}^n \binom{n}{k} \frac{(-1)^k}{(k+1)^2} = \frac{H_{n+1}}{n+1}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}

Note that
$$
\Gamma\pars{a + \mu} =
\Gamma\pars{a}\bracks{%
1  + \pars{H_{a - 1}\ -\ \gamma}\mu + \,\mrm{O}\pars{\mu^{2}}}
$$
such that
\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}
A: 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.
A: $(1-x)^n=\sum_{0\le k\le n}\binom nk(-1)^kx^k$
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.
A: An overkill. From the derivative version of the Melzak's identity $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{f\left(y-k\right)}{\left(x+k\right)^{2}}=\frac{f\left(x+y\right)\sum_{k=0}^{n}\frac{1}{k+x}-\frac{d}{dx}f\left(x+y\right)}{x\dbinom{x+n}{n}},\,x\neq-k$$ where $f$ is a polynomial up to degree $n$ we have, taking $f\left(z\right)\equiv1,\,x=1$, that $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{\left(1+k\right)^{2}}=\frac{\sum_{k=0}^{n}\frac{1}{k+1}}{n+1}=\color{red}{\frac{H_{n+1}}{n+1}}$$ as wanted.
A: I don't if you would accept this:
$$\sum_{k=0}^{n} \binom{n}{k}x^k=(1+x)^k$$
Integrate with respect to $x$ from $0$ to $t$ to get:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{t^{k+1}}{k+1}=\frac{(1+t)^{k+1}-1}{k+1}$$
Now divide by $t$ to get:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{t^{k}}{k+1}=\frac{(1+t)^{k+1}-1}{(k+1)t}$$
Now interate agian with respect to $t$ from $0$ to $z$ to get:
$$\sum_{k=0}^{n} \binom{n}{k}\frac{z^{k}}{(k+1)^2}=\int_{0}^z\frac{(1+t)^{k+1}-1}{(k+1)t}dt$$
Now Let $z=-1$
$$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^{k}}{(k+1)^2}=\int_{0}^{-1}\frac{(1+t)^{k+1}-1}{(k+1)t}dt$$
A: Suppose we seek to compute
$$S_n = \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+1)^2}.$$
With this in mind we introduce the function
$$f(z) = n! (-1)^n \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q}.$$
We then obtain for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z) =
(-1)^n \frac{n!}{(k+1)^2} \prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = (-1)^n \frac{n!}{(k+1)^2}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= {n\choose k} \frac{(-1)^k}{(k+1)^2}.$$
This means that
$$S_n = \sum_{k=0}^n \mathrm{Res}_{z=k} f(z)$$
and since residues sum to zero we have
$$S_n + \mathrm{Res}_{z=-1} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$
We can compute  the residue at infinity by inspection  (it is zero) or
more formally through
$$\mathrm{Res}_{z=\infty}
n! (-1)^n \frac{1}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q}
\\ = - n! (-1)^n \mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{1}{(1/z+1)^2} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = - n! (-1)^n \mathrm{Res}_{z=0}
\frac{1}{(z+1)^2} \prod_{q=0}^n \frac{z}{1-qz}
\\ = - n! (-1)^n \mathrm{Res}_{z=0}
\frac{z^{n+1}}{(z+1)^2} \prod_{q=0}^n \frac{1}{1-qz} = 0.$$
We get for the residue at $z=-1$ that
$$\mathrm{Res}_{z=-1} f(z) =
n! (-1)^n \left. \left(\prod_{q=0}^n \frac{1}{z-q}\right)'\right|_{z=-1}
\\ = - n! (-1)^n \left.
\left(\prod_{q=0}^n \frac{1}{z-q}\right)
\sum_{q=0}^n \frac{1}{z-q} \right|_{z=-1}
\\ = - n! (-1)^n \frac{(-1)^{n+1}}{(n+1)!} \left(-H_{n+1}\right)
= -\frac{H_{n+1}}{n+1}.$$
We thus have
$$S_n -\frac{H_{n+1}}{n+1} = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{
\frac{H_{n+1}}{n+1}.}$$
A: Let
$$
f(x)=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^kx^{k+1}}{(k+1)^2}
$$
then
$$
\begin{align}
f'(x)
&=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^kx^k}{(k+1)}\\
&=\frac1{n+1}\sum_{k=0}^n\binom{n+1}{k+1}(-1)^kx^k\\
&=\frac1{n+1}\frac{1-(1-x)^{n+1}}{x}
\end{align}
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^2}
&=f(1)\\
&=\frac1{n+1}\int_0^1\frac{1-(1-x)^{n+1}}{x}\,\mathrm{d}x\\
&=\frac1{n+1}\lim_{a\to1^-}\left[\int_0^1x^{-a}\,\mathrm{d}x-\int_0^1(1-x)^{n+1}x^{-a}\,\mathrm{d}x\right]\\
&=\frac1{n+1}\lim_{a\to1^-}\left[\frac1{1-a}-\frac{\Gamma(n+2)\Gamma(2-a)}{(1-a)\Gamma(n+3-a)}\right]\\
&=\frac1{n+1}\lim_{a\to1^-}\frac{1-(n+1)!\frac1{(n+2-a)(n+1-a)\cdots(2-a)}}{1-a}\\[3pt]
&=\frac1{(n+1)(n+1)!}\lim_{a\to1^-}\frac{(n+2-a)(n+1-a)\cdots(2-a)-(n+1)!}{1-a}\\[3pt]
&=\frac{H_{n+1}}{n+1}
\end{align}
$$
A: 
The answers have been given already, but just for seeing an elementary one I posted this answer:


$$\sum_{k=0}^{n}\binom{n}{k}\frac{\left(-1\right)^{k}}{\left(k+1\right)^{2}}=\frac{1}{n+1}\sum_{k=0}^{n}\binom{n+1}{k+1}\frac{\left(-1\right)^{k}}{k+1}$$
$$=\frac{-1}{n+1}\sum_{k=1}^{n+1}\binom{n+1}{k}\frac{\left(-1\right)^{k}}{k}$$$$=\frac{-1}{n+1}\left[\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{\left(-1\right)^{k}}{k}+\sum_{k=1}^{n+1}\binom{n}{k}\frac{\left(-1\right)^{k}}{k}\right]$$$$=\frac{-1}{n+1}\left[\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{\left(-1\right)^{k}}{k}+\sum_{k=1}^{n+1}\binom{n-1}{k-1}\frac{\left(-1\right)^{k}}{k}+\sum_{k=1}^{n+1}\binom{n-1}{k}\frac{\left(-1\right)^{k}}{k}\right]$$
On the other hand:
$$\sum_{k=1}^{n+1}\binom{n+1-r}{k-1}\frac{\left(-1\right)^{k}}{k}=\frac{1}{n+2-r}\sum_{k=1}^{n+1}\binom{n+2-r}{k}\left(-1\right)^{k}$$$$=\frac{1}{n+2-r}\left[\sum_{k=0}^{n+2-r}\binom{n+2-r}{k}\left(-1\right)^{k}+\sum_{k=n+3-r}^{n+1}\binom{n+2-r}{k}\left(-1\right)^{k}-1\right]$$$$=\frac{-1}{n+2-r}\tag{I}$$
Continuing this way:
$$\frac{-1}{n+1}\left[\sum_{k=1}^{n+1}\binom{n}{k-1}\frac{\left(-1\right)^{k}}{k}+...+\sum_{k=1}^{n+1}\binom{n+1-(n)}{k-1}\frac{\left(-1\right)^{k}}{k}+\color{blue}{\sum_{k=1}^{n+1}\binom{n+1-(n)}{k}\frac{\left(-1\right)^{k}}{k}}\right]$$
Using $\text{(I)}$ implies:
$$=\frac{-1}{n+1}\left[\sum_{k=0}^{n-1}\frac{-1}{n+1-k}\color{blue}{-1}\right]=\frac{1}{n+1}\sum_{k=1}^{n+1}\frac{1}{k}=\frac{H_{n+1}}{n+1}$$
Hence:
$$\bbox[5px,border:2px solid #00A000]{\sum_{k=0}^{n}\binom{n}{k}\frac{\left(-1\right)^{k}}{\left(k+1\right)^{2}}=\frac{H_{n+1}}{n+1}}$$
And the claim follows.
