# 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?

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Wolfram has a suggestion, which is beyond my comprehension. –  Hendrik Jan Dec 9 '12 at 10:26
The same answer with maple $\frac{\psi(n+2)+\gamma}{n+2}$. –  Mhenni Benghorbal Dec 9 '12 at 10:57
@Michael: can you add (as a one-liner) in what context you found the question? Just curious. –  Hendrik Jan Dec 9 '12 at 16:21

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.

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+1: This is definitely the best answer –  Amr Dec 9 '12 at 11:11
You are absolutely right Antonio, I just got to the same expression using the fact $\sum \binom{n}{k}\frac{(-1)^k}{x+k} = x^{-1}\binom{x+n}{n}^{-1}$ and taking the derivative on both sides. –  Michael Dec 9 '12 at 11:26
Just to confirm your answer with the others in the comment, here is the relation between the $\psi$ function and the harmonic numbers $H_n=\gamma+\psi_0(n+1)$. See here. –  Mhenni Benghorbal Dec 9 '12 at 14:21

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

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

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.

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you have missed all the binomial coefficients here –  dexter04 Dec 9 '12 at 11:00
The integration of $x^{k+1}/k+1$ is $x^{k+2}/(k+1)(k+2)$ –  Amr Dec 9 '12 at 11:03
Thanks, rectified the errors. –  lab bhattacharjee Dec 9 '12 at 11:30

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 $z=-1$ $$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^{k}}{(k+1)^2}=\int_{0}^{-1}\frac{(1+t)^{k+1}-1}{(k+1)t}dt$$

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