# Closed form for a binomial series

I am wondering if any knows how to compute a closed form for the following two series.

1. $$\sum_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m}$$
2. $$\sum_{m=1}^{n}\frac{(-1)^m}{m^4}\binom{2n}{n+m}$$

Mathematica gave a result in terms of the Hypergeometric function, which I don't understand. I was wondering if anyone can illustrate how to express these two series in a nice closed form, without using the Hypergeomeyric function. I was thinking along the line using a combinatorial identity or method, but I having problems figuring it out.

• Result in terms of the hypergeometric function is given when there are no nice closed forms , i.e., when there are no closed forms in terms of standard mathematical functions. – Prasun Biswas Jul 4 '15 at 3:19
• Somewhat nicer expression for (1) is $$2^{n-1}\left(\Psi^{(1)}(n+1)-\frac{\pi^2}{6}\right)\frac{(2n-1)!!}{n!}$$ – d.k.o. Jul 4 '15 at 3:51
• Thanks for the reply. This is an interesting formula. How did you go about deriving it? Do you have any insight on how to compute a nicer expression for the second identity? – Hatchet Jul 4 '15 at 6:29

Let $$g(n,m)=\frac1{\binom{2n}{n}}\sum_{j=1}^n\frac{(-1)^j}{j^m}\binom{2n}{n+j}\tag{1}$$ then it can be shown that $$g(n,m)=g(n-1,m)+\frac{g(n,m-2)}{n^2}\tag{2}$$ The proof of $(2)$ is below.

Since $$\sum_{j=0}^k(-1)^j\binom{n}{j}=(-1)^k\binom{n-1}{k}\tag{3}$$ we get $$g(n,0)=-\frac12\tag{4}$$ then, using $(2)$ and $(4)$, we get $$g(n,2)=-\frac12H_n^{(2)}\tag{5}$$ Therefore, $$\bbox[5px,border:2px solid #C0A000]{\sum_{j=1}^n\frac{(-1)^j}{j^2}\binom{2n}{n+j}=-\frac12H_n^{(2)}\binom{2n}{n}}\tag{6}$$ where $$H_n^{(k)}=\sum_{j=1}^n\frac1{j^k}\tag{7}$$

Using $(2)$ and $(5)$, we get that $$g(n,4)=-\frac14\left(\left(H_n^{(2)}\right)^2+H_n^{(4)}\right)\tag{8}$$ therefore, $$\bbox[5px,border:2px solid #C0A000]{\sum_{j=1}^n\frac{(-1)^j}{j^4}\binom{2n}{n+j}=-\frac14\left(\left(H_n^{(2)}\right)^2+H_n^{(4)}\right)\binom{2n}{n}}\tag{9}$$

Proof of $\boldsymbol{(2)}$:

Note that \begin{align} g(n,m) &=\sum_{j=1}^n\frac{(-1)^j}{j^m}\frac{n!}{(n-j)!}\frac{n!}{(n+j)!}\\ &=\frac{(-1)^n}{n^m\binom{2n}{n}}+\sum_{j=1}^{n-1}\frac{(-1)^j}{j^m}\frac{(n-1)!}{(n-j-1)!}\frac{(n-1)!}{(n+j-1)!}\frac{n^2}{n^2-j^2}\\ g(n-1,m) &=\hphantom{\frac{(-1)^n}{n^m\binom{2n}{n}}+}\sum_{j=1}^{n-1}\frac{(-1)^j}{j^m}\frac{(n-1)!}{(n-j-1)!}\frac{(n-1)!}{(n+j-1)!}\\ g(n,m)-g(n-1,m) &=\frac{(-1)^n}{n^m\binom{2n}{n}}+\sum_{j=1}^{n-1}\frac{(-1)^j}{j^m}\frac{(n-1)!}{(n-j-1)!}\frac{(n-1)!}{(n+j-1)!}\frac{j^2}{n^2-j^2}\\ &=\frac{(-1)^n}{n^m\binom{2n}{n}}+\frac1{n^2}\sum_{j=1}^{n-1}\frac{(-1)^j}{j^{m-2}}\frac{n!}{(n-j)!}\frac{n!}{(n+j)!}\\ &=\frac1{n^2}\sum_{j=1}^n\frac{(-1)^j}{j^{m-2}}\frac{n!}{(n-j)!}\frac{n!}{(n+j)!}\\ &=\frac{g(n,m-2)}{n^2} \end{align}

• Brilliant! Using a recursive relation was the ticket. Thanks for the help. – Hatchet Jul 5 '15 at 8:05

The first expression:

\begin{align}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m} &= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{n}{m}\frac{m!n!}{(m+n)!}\\ &= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m}\binom{n}{m}\frac{(m-1)!n!}{(m+n)!}\\ &= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m}\binom{n}{m}\int_0^1 x^{m-1}(1-x)^n\,dx\\ &= \binom{2n}{n}\int_0^1\int_0^{x}\sum\limits_{m=1}^{n}\binom{n}{m}(-1)^my^{m-1} \frac{(1-x)^n}{x}\,dy\,dx\\ &= \binom{2n}{n}\int_0^1\int_0^{x}\frac{(1-y)^n -1 }{y}. \frac{(1-x)^n}{x}\,dy\,dx\\ &= \binom{2n}{n}\int_0^1 \frac{(1-x)^n}{x}\int_0^{x}\frac{(1-y)^n -1 }{y} \,dy\,dx\\ &= \binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\int_0^{1-x}\frac{(1-y)^n -1 }{y} \,dy\,dx\\ &= -\binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\int_0^{1-x}\frac{1-(x+y)^n}{1-(x+y)} \,dy\,dx\\ &= -\binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\left[\sum\limits_{m=1}^{n}\frac{(x+y)^m}{m}\right]_0^{1-x}\,dx\\ &= -\binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\sum\limits_{m=1}^{n}\frac{1-x^m}{m}\,dx\\ &= -\binom{2n}{n}\sum\limits_{m=1}^{n}\frac{1}{m}\sum\limits_{k=1}^{m}\frac{1}{n+k}= -\binom{2n}{n}\left(\sum\limits_{m=1}^{n}\frac{H_{m+n}}{m} - H_n^2\right)\end{align}

The second one looks like a troublesome calculation to make.

$$\sum\limits_{m=1}^{n}\frac{(-1)^m}{m}\binom{2n}{n+m} = -\binom{2n}{n}(H_{2n} - H_n)$$

With @Chris's sis's help in chat I figured out how to simplify the expression further:

\begin{align}\sum\limits_{m=1}^{n}\frac{H_{m+n}}{m} &= \sum\limits_{m=1}^{n}\frac{H_m}{m}+\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{n} \frac{1}{m(k+m)}\\&= \frac{1}{2}(H_n^{(2)}+H_n^2) + \frac{1}{2}\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{n} \left(\frac{1}{m(k+m)}+\frac{1}{k(k+m)}\right)\\&= \frac{1}{2}(H_n^{(2)}+H_n^2) + \frac{1}{2}\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{n} \frac{1}{mk}\\&= H_n^2 + \frac{1}{2}H_n^{(2)} \tag{1} \end{align}

Hence, $\displaystyle \sum_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m} = -\frac{1}{2}\binom{2n}{n}H_n^{(2)}$

Edit: For the second one, a short computation similar to one above shows:

$$\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^4}\binom{2n}{n+m} = -\sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}$$

Couldn't figure out a more direct line of computation, used differencing to complete the proof:

Call the triple sum $\displaystyle F(n) = \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}$

Then,

\begin{align}F(n+1) - F(n) &= \sum\limits_{m=1}^{n+1}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n+1} - H_{n+1}}{mjk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}\\ &= \sum\limits_{m=1}^{n+1}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n - \frac{k}{(n+1)(n+k+1)} }{mjk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}\\ &= \frac{1}{n+1}\sum\limits_{j=1}^{n+1}\sum\limits_{k=1}^{j} \frac{H_{k+n+1} - H_{n+1}}{jk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{k}{mjk(n+1)(n+k+1)}\\ &= \frac{1}{n+1}\left(\sum\limits_{j=1}^{n+1}\sum\limits_{k=1}^{j} \frac{H_{k+n+1} - H_{n+1}}{jk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m} \frac{H_{n+j+1} - H_{n+1}}{mj}\right)\\&= \frac{1}{(n+1)^2}\sum\limits_{j=1}^{n+1} \frac{H_{n+j+1} - H_{n+1}}{j} = \frac{H_{n+1}^{(2)}}{2(n+1)^2}\end{align}

In the last line I used the previous result $(1)$.

Hence, $\displaystyle F(n) = \frac{1}{2}\sum\limits_{k=1}^{n} \frac{H_{k}^{(2)}}{k^2} = \frac{\left(H_{n}^{(2)}\right)^2 + H_{n}^{(4)}}{4}$

• Looks nice. Thanks. Do you know of any identities or ways to bound $(H_n^{(2)})^2$ – Hatchet Jul 5 '15 at 8:08
• @Hatchet I suppose we could use $\frac{3}{2} - \frac{1}{n+1} = 1+\sum\limits_{k=2}^{n} \frac{1}{k(k+1)} \le H_n^{(2)} = 1+\sum\limits_{k=2}^{n} \frac{1}{k^2} \le 1+\sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 2 - \frac{1}{n}$ as a weak bound. – r9m Jul 5 '15 at 8:36

This is not an answer but it is too long for a comment.

For the first expression, d.k.o. gave a very elegant expression involving the first derivative of the digamma function and I do not think that I could provide anything simpler. $$S_n=\sum_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m}=2^{n-1}\left(\Psi^{(1)}(n+1)-\frac{\pi^2}{6}\right)\frac{(2n-1)!!}{n!}$$ For large values of $n$, we can replace $\Psi^{(1)}(n+1)$ by the expansion $$\Psi^{(1)}(n+1)=\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{6 n^3}-\frac{1}{30 n^5}+O\left(\frac{1}{n^6}\right)$$ which leads to quite accurate results.

For example

• $S_{10}=-\frac{90915148181}{635040}\approx -143164.443$ while the above expansion gives $\approx -143164.413$.
• $S_{100}=\approx -7.40226818241923\times 10^{58}$ while the above expansion gives $\approx -7.40226818240414\times 10^{58}$.

For both expressions, we can notice that the logarithm of the negative value of the sum varies almost linearly with $n$.

• Thanks for the reply. Do you believe that both of these sums run parallel to a logarithm, i.e. can they be expressed as a logarithm? How would you find an elegant expression for the second series? – Hatchet Jul 4 '15 at 6:34