# Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$

Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $$H_n=\frac11+\frac12+\cdots+\frac1n$$?

Quick Results This series converges because $$\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}=O\left(\frac{\ln^2n}{n^{3/2}}\right).$$ My Attempt
Recall the integral representation of harmonic number $$H_n=\int_0^1\frac{1-x^n}{1-x}d x$$ we have $$S=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\iint_{[0,1]^2}\frac{(1-x^n)(1-y^n)}{(1-x)(1-y)}d xd y\\ =\tiny\iint_{[0,1]^2}\frac{x y \log (4)-2 x y \log \left(\sqrt{1-x}+1\right)-2 x y \log \left(\sqrt{1-y}+1\right)+2 x y \log \left(\frac{1}{2} \left(\sqrt{1-x y}+1\right)\right)}{\left(\sqrt{1-x y}-1\right) \left(\sqrt{1-x y}+1\right)}dxdy\\$$ This integral is too hard for me and Mathematica to compute. Numerical integration returns $$12.6178$$, it agrees with the numerical summation of the original series. I tried to integrate with respect to $$x$$, but failed.

• Have you any evidence that it has a closed form. Is this, maybe a contest or challenge problem? Apr 12, 2019 at 5:40
• @Lord I have found the closed-form expressions for many similar series, such as the series without the exponent 2. I believe, without evidence, that it has a closed form. Apr 12, 2019 at 5:49
• I think that the problem is because of the exponent $2$. I would enjoy seeing a closed form expression. Apr 12, 2019 at 6:24
• @Claude after cheating with Mathematica, I found a closed-form:$21\zeta(3)/2$. I will post a solution in my spare time. Apr 14, 2019 at 18:25
• Very nice problem. The funny thing is that 21/2 zeta(3) doesnt match the value of the sum numerically and that made me doubt my solution and spend too much time trying to spot any typo I might did. I wonder why wolfram gives the wrong numerical answer sometimes. Jun 7, 2019 at 6:27

First, we prove a lemma on the integral representation of $$(H_n)^2$$. $$I_n=\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)=(H_n)^2$$

Let's prove by induction. $$\displaystyle I_0=-\int_0^1\frac{\ln x}{1-x}dx=\zeta(2)=\zeta(2)+(H_0)^2$$.\ Assume the equation holds for $$n-1$$, \begin{aligned} I_n&=\int_0^1\left(2(x^n-1)\frac{\ln(1-x)}{1-x}-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\\ &=I_{n-1}+\int_0^1\left(2(x^n-x^{n-1})\frac{\ln(1-x)}{1-x}-\frac{(x^n-x^{n-1})\ln x}{1-x}\right)d x\\ &=(H_{n-1})^2+\int_0^1\left(-2x^{n-1}\ln(1-x)+x^{n-1}\ln x\right)d x\\ &=\left(H_n-\frac1n\right)^2-\frac1{n^2}+2\cdot\frac{H_n}n=(H_n)^2 \end{aligned} Result Therefore, and by integrating $$\displaystyle\sum_{n=1}^\infty\frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$$ from $$0$$ with respect to $$x$$, we have \begin{aligned} S&=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\left(\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\right)\\ &=\int_0^1\left(\frac{1}{x\sqrt{1-x}}-\frac1x\right)\ln^2(1-x)d x-\int_0^12\ln\frac{2}{1+\sqrt{1-x}}\frac{\ln x}{1-x}d x-2\ln2\zeta(2)\\ &=I_1-I_2-2\ln2\zeta(2) \end{aligned} $$I_1=12\zeta(3)$$ can be easily deduced by substitution $$x\mapsto 1-x^2$$. $$-2\ln2\zeta(2)+\frac32\zeta(3)$$, the value of $$I_2$$, can also be deduced by the same substitution. By combining these results, $$S=\frac{21}2\zeta(3)$$.

• Very nice, and with a surprisingly simple result. And it gives at the same time the closed expression of my integral (7). Forthermore it shows that my summation of the asymptitic summands (2) is misleading. Apr 15, 2019 at 17:59

we have $$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1 \quad$$ divide both sides by $$x$$ then integrate , we get

$$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+c$$
let $$x=0,\$$ we get $$C=2\ln2$$

then $$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=\color{orange}{-2 \tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2}$$

also we have $$\displaystyle -\int_0^1x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n} \tag{1}$$ multiply both sides by $$\displaystyle \frac{\binom{2n}n}{n4^n}\quad$$ then take the sum, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^2}\frac{\binom{2n}n}{4^n}&=-\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^ndx\\ &=-\int_0^1\frac{\ln(1-x)}{x}\left(\color{orange}{-2\tanh^{-1}{\sqrt{1-x}}-\ln x+2\ln2}\right)dx\\ &=\small{2\int_0^1\frac{\ln(1-x)\tanh^{-1}{\sqrt{1-x}}}{x}dx+\int_0^1\frac{\ln x\ln(1-x)}{x}dx-2\ln2\int_0^1\frac{\ln(1-x)}{x}dx}\\ &=2\int_0^1\frac{\ln x\tanh^{-1}{\sqrt{x}}}{1-x}dx+\int_0^1\frac{\ln x\ln(1-x)}{x}dx-2\ln2\int_0^1\frac{\ln x}{1-x}dx\\ &=8\int_0^1 \frac{x\ln x\tanh^{-1}x}{1-x^2}\ dx+\zeta(3)+2\ln2\zeta(2)\\ &=\color{blue}{8I+\zeta(3)+2\ln2\zeta(2)} \end{align}

differentiate $$(1)$$ w.r.t $$\ n\$$, we get $$\quad\displaystyle \int_0^1x^{n-1}\ln x\ln(1-x)dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}-\zeta(2)}{n}$$

multiply both sides by $$\ \displaystyle \frac{\binom{2n}n}{4^n}$$ then take the sum, we get, \begin{align} \sum_{n=1}^\infty \frac{H_n}{n^2}\frac{\binom{2n}n}{4^n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}-\zeta(2)\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}&=\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx\\ \color{blue}{8I+\zeta(3)+2\ln2\zeta(2)}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}-2\ln2\zeta(2)&=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\ &=\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}\ dx-\zeta(3)\\ &=\int_0^1\frac{\ln(1-x)\ln x}{(1-x)\sqrt{x}}\ dx-\zeta(3)\\ &=4\int_0^1\frac{\ln(1-x^2)\ln x}{1-x^2}\ dx-\zeta(3)\\ &=4K-\zeta(3) \end{align} rearranging the terms, we have $$\quad\displaystyle\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}=\color{red}{4K-8I-2\zeta(3)}$$

using the second derivative of beta function, we have $$\quad\displaystyle\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2}{n}+\frac{H_n^{(2)}}{n}$$

multiply both sides by $$\ \displaystyle \frac{\binom{2n}n}{4^n}$$ then take the sum, we get, \begin{align} \sum_{n=1}^\infty \frac{H_n^{2}}{n}\frac{\binom{2n}n}{4^n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n}\frac{\binom{2n}n}{4^n}&=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx\\ \sum_{n=1}^\infty \frac{H_n^{2}}{n}\frac{\binom{2n}n}{4^n}+\color{red}{4K-8I-2\zeta(3)}&=\int_0^1\frac{\ln^2(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\ &=\int_0^1\frac{\ln^2(1-x)}{x\sqrt{1-x}}\ dx-\int_0^1\frac{\ln^2(1-x)}{x}\ dx\\ &=\int_0^1\frac{\ln^2x}{(1-x)\sqrt{x}}\ dx-\int_0^1\frac{\ln^2x}{1-x}\ dx\\ &=8\int_0^1\frac{\ln^2x}{1-x^2}\ dx-2\zeta(3)\\ &=8\left(\frac74\zeta(3)\right)-2\zeta(3)\\ &=12\zeta(3) \end{align} rearranging the terms, we have $$\quad\displaystyle\sum_{n=1}^\infty \frac{H_n^2}{n}\frac{\binom{2n}n}{4^n}=14\zeta(3)+8I-4K$$

by applying IBP for$$\ I$$, we get $$\quad\displaystyle8I=4K+4\int_0^1\frac{\ln(1-x^2)\tanh^{-1}x}{x}dx$$

then \begin{align} \sum_{n=1}^\infty\frac{H_n^2}{n}\frac{\binom{2n}n}{4^n}&=14\zeta(3)+4\int_0^1\frac{\ln(1-x^2)\tanh^{-1}x}{x}dx\\ &=14\zeta(3)+2\int_0^1\frac{\left[\ln(1+x)+\ln(1-x)\right]\left[\ln(1+x)-\ln(1-x)\right]}{x}dx\\ &=14\zeta(3)+2\int_0^1\frac{\ln^2(1+x)-\ln^2(1-x)}{x}dx\\ &=14\zeta(3)+2\left(\frac14\zeta(3)-2\zeta(3)\right)\\ &=\frac{21}{2}\zeta(3) \end{align}

This is not a complete solution but some first steps.

EDIT 12.04.19 23:20

Much simpler single integral derived.

Original post

The sum in question is

$$s = \sum_{n=1}^\infty a_n\tag{1}$$

with

$$a_n = \frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}\tag{2}$$

1. Representation as a single integral

1.1

Let us replace just one harmonic number in $$a_n$$.

Using the definition

$$H_n = \sum _{k=1}^{\infty } \frac{n}{k (k+n)}\tag{3}$$

and writing

$$\frac{1}{n+k}=\int_0^1 x^{n+k-1}\,dx\tag{4}$$

gives for the n-sum

$$\sum_{n=1}^{\infty } \frac{\binom{2 n}{n} H_n x^n}{4^n}=\frac{\partial}{\partial{c}} \left( {_2}F{_1} \left( \frac{1}{2},1,c,x\right)\right)|_{ c \to 1}\tag{5}$$

The remaining k-sum is easily done

$$-\sum _{k=1}^{\infty } \frac{x^{k-1}}{k} =\frac{\log (1-x)}{x}$$

Hence $$s$$ can be expressed as

$$s_1 = \frac{\partial}{\partial{c}} i(c)|_{ c \to 1} \tag{6a}$$

with

$$i(c) = \int_0^1 \frac{\log (1-x)}{x} {_2}F{_1} \left( \frac{1}{2},1,c,x\right)\,dx\tag{6b}$$

Here $${_2}F{_1}$$ is the hypergeometric function.

Numerically, we find in this form

$$s = 12.6216...$$.

1.2 Simpler single integral

The expression derived in the previous paragraph is correct but not very useful because it contains the hypergeometric function. Here we derive the following simpler formula with an elementary integrand.

$$s_2 = \int_0^\infty \frac{v}{\sinh \left(\frac{v}{2}\right)} \left(\frac{v}{\sqrt{2-e^{-v}}}-2 \log \left(\frac{\sqrt{2-e^{-v}}+1}{e^{-\frac{v}{2}}+1}\right)\right)\,dv\tag{7}$$

This is a well converging integral, suited for numerical evaluation. The integrand is depicted here

The derivation starts with replacing both $$H_n$$ by (3) and (4).

This gives the integral

$$s = \int_0^1 \int_0^1 \frac{\log(1-x) \log(1-y)}{2(1-x y )^{\frac{3}{2}}}\,dx\,dy\tag{8}$$

Transforming $$x\to 1-e^{-u}$$, $$y\to 1-e^{-v}$$ leads to

$$s = \int_0^\infty \int_0^v (u v ) \frac{e^{\frac{u+v}{2}}}{(e^u + e^v -1 )^{\frac{3}{2}}}\,du\,dv\tag{8}$$

Here we have made use of the symmetry of the integrand to restrict the integration region to $$u\le v$$ (and applying a factor 2). Luckily the $$u$$-integral can be done with the result (7).

2. Sum with asymptotic summands

An attempt to get a feeling for the ingredients of a possible closed form.

The leading asymptotic term of $$a_n$$ is

$$a_n \simeq b_n = \frac{(\log (n)+\gamma )^2}{\sqrt{\pi } n^{3/2}}\tag{1}$$

The sum of $$b_n$$ instead of $$a_n$$ gives

$$s \simeq \sum_{n=1}^\infty b_n = \frac{1}{\sqrt{\pi }}\left(\zeta ''\left(\frac{3}{2}\right)-2 \gamma \zeta '\left(\frac{3}{2}\right)+\gamma ^2 \zeta \left(\frac{3}{2}\right)\right)\simeq 12.0733\tag{2}$$

Here $$\zeta(x)$$ is the Riemann zeta function (and it derivatives), and $$\gamma$$ is the Euler-Mascheroni constant.

Notice that the numerical value is close to the one mentioned in the OP. Taking higher terms in the asymptotic expansion of $$a_n$$ leads to slightly higher numerical values.

A similar approach to my previous solution above but neater:

Using the well-known identity

$$\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$$

Divide both sides by $$x$$ then integrate , we get

$$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+C$$
set $$x=0,\$$ we get $$C=2\ln2$$

Then

$$\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+2\ln2\tag1$$

Multiply both sides of (1) by $$-\frac{\ln(1-x)}{x}$$ then integrate from $$x=0$$ to $$1$$ and use the fact that $$-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$$ we get

\begin{align} \sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}&=2\underbrace{\int_0^1\frac{\ln(1+\sqrt{1-x})\ln(1-x)}{x}dx}_{\sqrt{1-x}=y}-2\ln2\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{-\zeta(2)}\\ &=8\int_0^1\frac{y\ln(1+y)\ln y}{1-y^2}dy+2\ln2\zeta(2)\\ &=4\int_0^1\frac{\ln(1+y)\ln y}{1-y}-4\int_0^1\frac{\ln(1+y)\ln y}{1+y}+2\ln2\zeta(2) \end{align}

where the first integral is

$$\int_0^1\frac{\ln y\ln(1+y)}{1-y}\ dy=\zeta(3)-\frac32\ln2\zeta(2)$$

and the second integral is

$$\int_0^1\frac{\ln y\ln(1+y)}{1+y}\ dy=-\frac12\int_0^1\frac{\ln^2(1+y)}{y}dy=-\frac18\zeta(3)$$

Combine the results of the two integrals we get

$$\boxed{\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)}$$

If we differentiate both sides of $$\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$$ we get

$$\int_0^1x^{n-1}\ln x\ln(1-x)dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}-\zeta(2)}{n}\tag2$$

Now multiply both sides of $$(2)$$ by $$\frac{1}{4^n}{2n\choose n}$$ the sum up from $$n=1$$ to $$\infty$$ we get

$$\sum_{n=1}^\infty \frac{H_n}{n^24^n}{2n\choose n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}-\zeta(2)\sum_{n=1}^\infty \frac{1}{n4^n}{2n\choose n}\\=\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\=\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx}_{\text{Beta function:}7\zeta(3)-6\ln2\zeta(2)}-\underbrace{\int_0^1\frac{\ln x\ln(1-x)}{x}dx}_{\zeta(3)}$$

Substitute $$\sum_{n=1}^\infty\frac{H_n}{n^24^n}{2n\choose n}=\frac92\zeta(3)-4\ln2\zeta(2)$$ and $$\sum_{n=1}^\infty\frac{1}{n4^n}{2n\choose n}=2\ln2$$ we get

$$\boxed{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n4^n}{2n\choose n}=\frac32\zeta(3)}$$

Using the identity

$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}\tag3$$

Again multiply both sides of $$(3)$$ by $$\frac{1}{4^n}{2n\choose n}$$ the sum up from $$n=1$$ to $$\infty$$ we get

$$\sum_{n=1}^\infty \frac{H_n^{2}}{n4^n}{2n\choose n}+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}\\=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n\ dx=\int_0^1\frac{\ln^2(1-x)}{x}\left(\frac{1}{\sqrt{1-x}}-1\right)\ dx\\=\underbrace{\int_0^1\frac{\ln^2(1-x)}{x\sqrt{1-x}}dx}_{\text{Beta function:}14\zeta(3)}-\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}dx}_{2\zeta(3)}$$

Finally, substitute $$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n4^n}{2n\choose n}=\frac32\zeta(3)$$ we get

$$\boxed{\sum_{n=1}^\infty \frac{H_n^{2}}{n4^n}{2n\choose n}=\frac{21}2\zeta(3)}$$

Addendum: Above, We calculated two integrals using Beta function but we can do them in a different way:

For the first integral $$\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx=\int_0^1\frac{\ln(1-x)\ln x}{(1-x)\sqrt{x}}dx=\int_0^1\frac{x^{-1/2}\ln x\ln(1-x)}{1-x}dx$$

we can use the generalization

$$\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right)$$

set $$m=1$$ then let $$n$$ approach $$-1/2$$ we get

$$\int_0^1\frac{\ln x\ln(1-x)}{x\sqrt{1-x}}dx=7\zeta(3)-6\ln2 \zeta(2)$$

For the second integral, set $$\sqrt{1-x}=y$$

$$\int_0^1\frac{\ln^2(1-x)}{x\sqrt{1-x}}dx=8\int_0^1\frac{\ln^2y}{1-y^2}dy\\=8\sum_{n=0}^\infty\int_0^1 x^{2n}\ln^2y\ dy=16\sum_{n=0}^\infty\frac{1}{(2n+1)^3}=16\left(\frac{7}{8}\zeta(3)\right)=14\zeta(3)$$

A Little bonus:

Multiply both sides of $$\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$$ by $$-\frac{\ln(1-x)}{x}$$ then integrate from $$x=0$$ to $$1$$ and use the fact that $$-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$$ we get

$$\sum_{n=1}^\infty\frac{H_n}{n4^n}{2n\choose n}=-\underbrace{\int_0^1\frac{\ln(1-x)}{x\sqrt{1-x}}dx}_{\sqrt{1-x}=y}+\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{-\zeta(2)}\\=-4\int_0^1\frac{\ln y}{1-y^2}dy-\zeta(2)=-4\left(-\frac34\zeta(2)\right)-\zeta(2)=\boxed{2\zeta(2)}$$

From this paper, Eq $$(13)$$ page 4, we have

$$\sum_{n=1}^\infty H_n\binom{2n}n x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right).$$

Replace $$x$$ by $$\frac{x}{4}$$, multiply both sides by $$-\frac{\ln(1-x)}{x}$$ then integrate using $$-\int_0^1 x^{n-1} \ln(1-x)dx=\frac{H_n}{n}$$, we have

$$\sum_{n=1}^\infty \frac{H_n^2}{n4^n}\binom{2n}n=-\int_0^1\frac{\ln(1-x)}{x\sqrt{1-x}}\ln\left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)dx$$

$$\overset{\sqrt{1-x}=y}{=}-8\int_0^1\frac{\ln y}{1-y^2}\ln\left(\frac{1+y}{2y}\right)dy$$

$$\overset{y=\frac{1-x}{1+x}}{=}4\int_0^1\frac{\ln^2(1-x)}{x}dx-4\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$

$$=4[2\zeta(3)]-4\left[-\frac58\zeta(3)\right]=\frac{21}{2}\zeta(3).$$