# Evaluate $\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$

By accident, I find this summation when I pursue the particular value of $$-\operatorname{Li_2}(\tfrac1{2})$$, which equals to integral $$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$$.

Notice this observation

$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x} = \int_{0}^{1} {\frac{\ln(1-x^{2})}{1+x} \mathrm{d}x} - \frac{(\ln2)^{2}}{2}$$

And using the Taylor series of $$\ln(1-x^{2})$$, I found this summation $$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$$, where $$H_{n}$$ is the harmonic-numbers.

If the value of $$\operatorname{Li_2}(\tfrac1{2})=\tfrac1{2}(\zeta(2)-(\ln2)^{2})$$ is given, the result can be easily deduced, which is

$$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)} = -\frac{\zeta(2)}{2}+(\ln2)^{2}$$

For the original goal is to calculate $$\operatorname{Li_2}(\tfrac1{2})$$, I expect some other approaches to the summation without using the value of $$\operatorname{Li_2}(\tfrac1{2})$$. I already knew some famous problem like Euler's Sum, which holds very similar form to this summation, but still in trouble finding the appropriate path.

## 2 Answers

Well, ignoring the dilogarithm reflection formula, we still have $$\sum_{n=1}^{N}\frac{\log(2)}{n}=\log(2)H_N,\qquad \sum_{n=1}^{N}\frac{H_n}{n}\stackrel{\text{sym}}{=}\frac{H_n^2+H_n^{(2)}}{2}$$ and $$\sum_{n=1}^{N}\frac{H_{2n}}{n}\stackrel{\text{SBP}}{=}H_N H_{2N}-\sum_{n=1}^{N-1}H_n\left(\frac{1}{2n+2}+\frac{1}{2n+1}\right)$$ can be reduced (up to known terms) to $$\sum_{n=1}^{N}\left[\frac{1}{n}\sum_{k=1}^{n}\frac{1}{2k-1}+\frac{1}{2n-1}\sum_{k=1}^{n}\frac{1}{k}\right]=\sum_{n=1}^{N}\frac{1}{n}\sum_{n=1}^{N}\frac{1}{2n-1}+\sum_{n=1}^{N}\frac{1}{n(2n-1)}.$$ Exploiting $$H_n^{(2)}=\zeta(2)+o(1)$$ and $$H_n = \log(n)+\gamma+o(1)$$ for $$n\to +\infty$$ we end up with the explicit value of $$\text{Li}_2\left(\frac{1}{2}\right)$$. Nice exercise!

• thanks for helpful answer! – Nanayajitzuki Mar 27 at 13:21

With this answer I show an indirect method to the wished result of the integral $$~\int\limits_0^1\frac{\ln(1-x)}{1+x}dx~$$,

and indirect means here: It’s used $$~\text{Li}_2\left(\frac{1}{2}\right)~$$ without knowing it’s value, only as a catalyst.

First a general formula. It’s not difficult to find out, that formally holds:

$$-\frac{d}{dx}(x+z)^y \sum\limits_{k=1}^\infty\frac{\left(\frac{x+z}{a+z}\right)^k}{k+y} = \frac{(x+z)^y}{x-a}$$

With the integration to $$x$$ and using Taylor series for $$~y~$$ around $$~0~$$ we get:

$$\frac{1}{n!}\int\frac{(\ln(x+z))^n}{x-a}dx = \sum\limits_{k=0}^n (-1)^{n-k+1}\frac{(\ln(x+z))^k}{k!}\text{Li}_{n-k+1}\left(\frac{x+z}{a+z}\right) + C$$

First we use $$~(n;z;a):=(1;1;1)~$$ :

$$\displaystyle\int\limits_0^1\frac{\ln(1-x)}{1+x}dx = -\int\limits_{-1}^0\frac{\ln(1+x)}{1-x}dx =$$

$$\displaystyle = -\text{Li}_2\left(\frac{x+1}{2}\right)|_{-1}^0 + \ln(x+1)\text{Li}_1\left(\frac{x+1}{2}\right)|_{-1}^0 = -\text{Li}_2\left(\frac{1}{2}\right)$$

Our next step is to transform the integral by partial integration :

$$\displaystyle\int\limits_0^1\frac{\ln(1-x)}{1+x}dx = (\ln(1-x))(\ln(1+x) - \ln 2)|_0^1 + \int\limits_0^1\frac{\ln(1+x) - \ln 2}{1-x}dx =$$

$$\displaystyle = 0 + 2\int\limits_0^{1/2}\frac{\ln(1+2x) - \ln 2}{1-2x}dx = -\int\limits_0^{1/2}\frac{\ln(x+1/2)}{x-1/2}dx$$

Now we use $$~(n;z;a):=(1;\frac{1}{2};\frac{1}{2})~$$ and $$~\text{Li}_1\left(\frac{1}{2}\right)=\ln 2~$$ :

$$\displaystyle -\int\limits_0^{1/2}\frac{\ln(x+1/2)}{x-1/2}dx = -\text{Li}_2\left(x+\frac{1}{2}\right)|_0^{1/2} + \ln\left(x+\frac{1}{2}\right) \text{Li}_1\left(x+\frac{1}{2}\right)|_0^{1/2}$$

$$\displaystyle = -\frac{\pi^2}{6} + \text{Li}_2\left(\frac{1}{2}\right) + (\ln 2)^2 \enspace\enspace$$ which is, as found before, the same as $$~\displaystyle -\text{Li}_2\left(\frac{1}{2}\right)~$$ .

Comparing both results we get the wished formula.

Note: $$~$$ Here we see very well that the partial integration leads to the second representation of the result and both representations have as a common base the (yellow marked) general formula.

Hint:

$$\frac{1}{n!}\int\frac{(\ln(x+z))^n}{(x-a)^{m+1}}dx =\\ =\frac{(-1)^m}{m!(a+z)^m}\sum\limits_{k=0}^n (-1)^{n-k+1}\frac{(\ln(x+z))^k}{k!}\sum\limits_{j=0}^m\begin{bmatrix}m~\\j~\end{bmatrix}\text{Li}_{n-k+1-j}\left(\frac{x+z}{a+z}\right) + C$$

for $$~m\in\mathbb{N}_0~$$ and with

the Stirling numbers of the first kind $$\begin{bmatrix}n~\\k~\end{bmatrix}~$$ defined by $$~\displaystyle\sum\limits_{k=0}^n\begin{bmatrix}n~\\k~\end{bmatrix}x^k:=\prod\limits_{k=0}^{n-1}(x+k)~$$

A simple example: $$\enspace\displaystyle\int\limits_0^1\frac{\ln(1-x)}{(1+x)^3}dx = \displaystyle -\int\limits_{-1}^0\frac{\ln(x+1)}{(x-1)^3}dx =-\frac{1+\ln 2}{8}$$

• so your path is actually to solve both the summation and $\text{Li}_2(\tfrac1{2})$ from its equivalent integral, it's acceptable for me, and your general formula really helpful for my further calculation, upvote. – Nanayajitzuki Mar 29 at 3:35
• @Nanayajitzuki : I've written only about the integral because you've found the summation yourself, so that it wasn't necessary to write anything more about that. ;) With the general formula I want to show the connection of both solutions (that they have the same base). – user90369 Mar 29 at 6:52