# 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.

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!

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. Mar 29 '19 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). Mar 29 '19 at 6:52

Here is an approach that avoids knowing the value of $$\operatorname{Li}_2 (\frac{1}{2})$$.

Let $$S = \sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ).$$ Observing that $$\int_0^1 \frac{x^{2n}}{1 + x} \, dx = H_n - H_{2n} + \ln 2,$$ your sum can be rewritten as \begin{align} S &= -\int_0^1 \frac{1}{1 + x} \sum_{n = 1}^\infty \frac{x^{2n}}{n} \, dx\\ &= \int_0^1 \frac{\ln (1 - x^2)}{1 + x} \, dx\\ &= \int_0^1 \frac{\ln (1 + x)}{1 + x} \, dx + \int_0^1 \frac{\ln (1 - x)}{1 + x} \, dx\\ &= I + J. \end{align} For the first of the integrals $$I$$, one has $$I = \frac{1}{2} \ln^2 2.$$ Now consider $$J - I$$. Then $$J - I = \int_0^1 \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{1 + x}.$$ Employing a self-similar substitution of $$t = (1-x)/(1+x)$$ leads to \begin{align} J - I &= \int_0^1 \frac{\ln t}{1 + t} \, dt\\ &= \sum_{n = 0}^\infty (-1)^n \int_0^1 t^n \ln t \, dt\\ &= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\int_0^1 t^{n + s} \, dt \right ]_{s = 0}\\ &= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\ &= -\underbrace{\sum_{n = 0}^\infty \frac{(-1)^n}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\ &= -\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^2}\\ &= -\sum_{n = 1}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= -\frac{1}{2} \zeta (2). \end{align} Thus $$J = I - \frac{1}{2} \zeta (2) = \frac{1}{2} \ln^2 2 - \frac{1}{2} \zeta (2).$$ Since $$S = I + J$$, we immediately see that $$\sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ) = \ln^2 2 - \frac{1}{2} \zeta (2),$$ as desired.

• Or we can do $J$ as follows$$J=\int_0^1\frac{\ln(1-x)}{1+x}\ dx=\int_0^1\frac{\ln x}{2-x}\ dx=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln x\ dx\\=-\sum_{n=1}^\infty\frac{1}{n^22^n}=-\operatorname{Li}_2(1/2)=\frac12\ln^22-\frac12\zeta(2)$$ Oct 23 '19 at 4:37
• @Ali Shather - Indeed, but the approached I used was designed to avoid using the known value for $\operatorname{Li}_2 \left (\frac{1}{2} \right )$. Oct 23 '19 at 4:41
• ah sorry .. you did it nicely by the way. Oct 23 '19 at 4:50