Sum $\sum\limits_{n=1}^\infty\frac{H_n^2}{n^22^n}$

Where $$H_n$$ is the harmonic number, $$\ \displaystyle H_n=1+\frac12+\frac13+...+\frac1n$$.

I am going to present my solution as I need it as a reference.

Other approaches are appreciated.

here is the closed form $$\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}=-\frac1{24}\ln^42+\frac14\ln^22\zeta(2)-\frac74\ln2\zeta(3)+\frac{37}{16}\zeta(4)-\operatorname{Li}_4\left(\frac12\right)$$

• My hunch is that there is no closed form for such a series. Have you tried anything? It's customary to show some effort... – Luke Collins Jun 9 at 22:46
• Do you have an integral representation for it to share? (since you put the integration tag). – カカロット Jun 9 at 22:54
• Are you looking for a value/closed-form, or for a proof the series converges? – Clement C. Jun 9 at 23:07
• @LukeCollins already solved :) – Ali Shather Jun 9 at 23:15
• @ Zacky its put in integral representation in the solution. – Ali Shather Jun 9 at 23:16

The problem reduces to the main integral in OP's post (see his answer).

Here's an alternative approach. Consider the following integrals: $$I=\int_0^1 \frac{\ln^2 x\ln(1+x)}{1-x}dx$$ $$J=\int_0^1 \frac{\ln^2 x\ln(1-x)}{1-x}dx=-\frac12\zeta(4)$$

$$X=I+J=\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1-x}dx$$ $$Y=\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1+x}dx$$ $$X+Y=2\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1-x^2}dx$$ $$X-Y=2\int_0^1 \frac{x\ln^2 x\ln(1-x^2)}{1-x^2}dx\overset{x^2\to x}=\frac14J$$

$$\Rightarrow I = \underbrace{\frac12\left((X+Y)+(X-Y)\right)}_{=X=I+J}-J=\frac12 (X+Y)-\frac78 J$$ $$=\int_0^1 \frac{\ln^2 x\ln(1-x^2)}{1-x^2}dx+\frac7{16}\zeta(4)$$ $$=\frac72\ln 2 \zeta(3)-\frac{45}{16}\zeta(4)+\frac7{16}\zeta(4)=\boxed{\frac72\ln 2 \zeta(3)-\frac{19}{8}\zeta(4)}$$

The last integral can be computed either using Beta function or with the power series $$\frac{\ln(1-x^2)}{1-x^2}=-\sum_{n=1}^\infty H_n x^{2n}$$

The second derivative of beta function gives $$\ \displaystyle \int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2}{n}+\frac{H_n^{(2)}}{n}$$

divide both sides by $$\ n2^n$$ and take the sum, we get \begin{align} \sum_{n=1}^\infty\frac{H_n^2}{n^22^n}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}&=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{(x/2)^n}{n} dx=-\int_0^1\frac{\ln^2(1-x)\ln(1-x/2)}{x} dx\\ &=-\int_0^1\frac{\ln^2(1-x)\left[\ln(2-x)-\ln2\right]}{x}\ dx, \quad 1-x=y\\ &=\ln2\int_0^1\frac{\ln^2x}{1-x} dx-\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\ dx\\ &=2\ln2\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^n\ln^2x}{1-x}\ dx\\ &=2\ln2\zeta(3)+\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(2\zeta(3)-2H_n^{(3)}\right)\\ &=2\ln2\zeta(3)-2\ln2\zeta(3)-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}\\ &=-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n} \end{align} then $$\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}=-\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}$$ the first sum can be found here, as for the second one, can be calculated as follows:

using the generating function $$\displaystyle\sum_{n=1}^\infty z^nH_n^{(3)}=\frac{\operatorname{Li}_3(z)}{1-z}$$, divide both sides by $$z$$ and integrate from $$z=0$$ to $$x$$,

then $$\quad\displaystyle\sum_{n=1}^\infty\frac{x^nH_n^{(3)}}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)\$$ and by taking $$x=-1$$, we get $$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$ plugging the closed forms of these two sums, we get

$$\sum_{n=1}^\infty\frac{H_n^2}{n^22^n}=-\frac1{24}\ln^42+\frac14\ln^22\zeta(2)-\frac74\ln2\zeta(3)+\frac{37}{16}\zeta(4)-\operatorname{Li}_4\left(\frac12\right)$$

• Basically, the only problem was $\int_0^1 \frac{\ln^2 x\ln(1+x)}{1-x}dx$ right? – カカロット Jun 10 at 0:06
• @Zacky yes but the real problem is choosing the right path that leads us to such managable integral. – Ali Shather Jun 10 at 0:09

Right here possibly the simplest way of dealing with the integral of the type $$\displaystyle I=\int_0^1 \frac{\ln^2 x\ln(1+x)}{1-x}\textrm{d}x$$ and its generalization.