# A closed form of $\sum_{n=1}^{\infty}(-1)^{n}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)$?

I’m curious about a possible closed form of the following series.

$$\sum_{n=1}^{\infty}(-1)^{n}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right) \tag1$$

One may observe that $(1)$ is absolutely convergent.

One may notice that apparently one can’t apply the same route that proved
$$\sum_{n=1}^{\infty}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)=\frac12\ln^2 2. \tag2$$

#### Solutions Collecting From Web of "A closed form of $\sum_{n=1}^{\infty}(-1)^{n}\ln \!\left(1+\frac1{2n}\right) \!\ln\!\left(1+\frac1{2n+1}\right)$?"

By Frullani’s theorem the given series can be represented as

$$\sum_{n\geq 1}(-1)^n \iint_{(0,+\infty)^2}\frac{\left(e^{-2nx}-e^{-(2n+1)x}\right)\left(e^{-(2n+1)y}-e^{-(2n+2)y}\right)}{xy}\,dx\,dy$$
that equals
$$-\int_{0}^{+\infty}\int_{0}^{+\infty}\frac{(e^x-1)(e^y-1)e^{-(2x+y)}}{xy(e^{2x+2y}+1)}\,dx\,dy$$
and maybe such double integral can be simplified by exploiting the properties of the dilogarithm function. Is is worth noticing that the previous representation is enough for providing an accurate numerical evaluation, $\approx -0.089$.

By following Pranav Arora’s approach to the other question, the given series equals
$$\sum_{n\geq 1}\log^2\left(1+\frac{1}{4n-2}\right)-\sum_{n\geq 1}\log^2\left(1+\frac{1}{4n-3}\right)+\frac{1}{2}\log^2(2)$$
hence the problem boils down to finding a closed form for
$$\sum_{n\geq 1}\log^2\left(1+\frac{1}{4n-k}\right),\qquad k\in\{0,1,2,3\}.$$

This has been done by the OP in this question, Theorem $(3)$. This kind of series can be written in terms of poly-Hurwitz zeta functions. By exploiting the Taylor series of $\log^2(1-z)$ and the integral representation for the harmonic numbers we have, for instance:
$$\begin{eqnarray*}\sum_{n\geq 1}\log^2\left(1+\frac{1}{n}\right)=\color{blue}{-2\int_{0}^{1}\frac{\log\Gamma(1+z)}{z(1-z)}\,dz} &=& 2 \int_{0}^{1}\psi(z+1)\log\left(\frac{z}{1-z}\right)\,dz\\&=&2\int_{0}^{1}\left(\frac{1}{z}-\frac{1}{1-z}-\pi \cot(\pi z)\right)\log(z)\,dz\\&=&\int_{0}^{1}2\text{arctanh}(t)\left(\frac{4t}{t^2-1}+\pi\tan\left(\frac{\pi t}{2}\right)\right)\,dt \end{eqnarray*}$$
where the blue integral can be further manipulated through Binet’s $\log\Gamma$ formulas.
The last approach is an idea of Cornel Ioan Valean: I am really grateful to him for the suggestion.

Since $\left(\frac{1}{z}-\frac{1}{1-z}-\pi \cot(\pi z)\right)$ is horribly close to $2z-1$ on the interval $(0,1)$, we have for instance $\sum_{n\geq 1}\log^2\left(1+\frac{1}{n}\right)<1$, and such inequality is pretty tight. Another idea from Cornel Ioan Valean is the following one: due to the Taylor series of the $\pi z\cot(\pi z)$ function,

$$\sum_{n\geq 1}\log^2\left(1+\frac{1}{n}\right)=2\zeta(2)-\sum_{n\geq 1}\frac{\zeta(2n)}{n^2}=\text{Li}_2(1)-\sum_{n\geq 2}\text{Li}_2\left(\frac{1}{n^2}\right)$$
and in a similar way the series $\sum_{n\geq 1}\log^2\left(1+\frac{1}{na+b}\right)$ can be written in terms of modified $L$-series like $\sum_{n\geq 1}\frac{\chi(n)\,\zeta(2n)}{n^2}$. The previous identity can also be seen as a consequence of the dilogarithm reflection formulas

$$\text{Li}_2(1-x)+\text{Li}_2(1-x^{-1})=-\frac{1}{2}\log^2 x\qquad\text{and}\qquad \text{Li}_2(x)+\text{Li}_2(-x)=\frac{1}{2}\text{Li}_2(x^2).$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

As an explicit petition of user $\color{#88f}{\texttt{@Brevan Ellefsen}}$ I wrote ( I guess so ) a set of hints. I hope it’ll be useful.

\begin{align}
&\sum_{n = 1}^{\infty}\pars{-1}^{n}\,\ln\pars{1 + {1 \over 2n}}
\ln\pars{1 + {1 \over 2n + 1}} =
\sum_{n = 1}^{\infty}\ic^{2n}\,\ln\pars{1 + {1 \over 2n}}
\ln\pars{1 + {1 \over 2n + 1}}
\\[5mm] = &\
\sum_{n = 2}^{\infty}\ic^{n}\,\ln\pars{1 + {1 \over n}}
\ln\pars{1 + {1 \over n + 1}}\,{1 + \pars{-1}^{n} \over 2} =
\Re\sum_{n = 1}^{\infty}\ic^{n}\,\ln\pars{1 + {1 \over n}}
\ln\pars{1 + {1 \over n + 1}}
\\[1cm] = &\
\phantom{-\,\,\,}{1 \over 2}\,\Re\sum_{n = 1}^{\infty}\ic^{n}\,
\ln^{2}\pars{\bracks{1 + {1 \over n}}\bracks{1 + {1 \over n + 1}}}
\\[2mm] &\ –
{1 \over 2}\,\Re\sum_{n = 1}^{\infty}\ic^{n}\,\ln^{2}\pars{1 + {1 \over n}} –
{1 \over 2}\,\Re\sum_{n = 1}^{\infty}\ic^{n}\,\ln^{2}\pars{1 + {1 \over n + 1}}
\\[1cm] = &\
{1 \over 2}\,\Re\sum_{n = 1}^{\infty}\ic^{n}\,
\ln^{2}\pars{1 + {2 \over n}} –
{1 \over 2}\,\Re\sum_{n = 1}^{\infty}\ic^{n}\,\ln^{2}\pars{1 + {1 \over n}} –
{1 \over 2}\,\Re\sum_{n = 2}^{\infty}\ic^{n – 1}\,\ln^{2}\pars{1 + {1 \over n}}
\\[1cm] = &\
{1 \over 2}\sum_{n = 1}^{\infty}\pars{-1}^{n}\,
\ln^{2}\pars{1 + {1 \over n}} –
{1 \over 2}\,\Re\sum_{n = 1}^{\infty}\ic^{n}\,\ln^{2}\pars{1 + {1 \over n}} –
{1 \over 2}\,\Im\sum_{n = 1}^{\infty}\ic^{n}\,\ln^{2}\pars{1 + {1 \over n}}
\\[2mm] + & {1 \over 2}\,\ln^{2}\pars{2}
\\ = &\
\mrm{f}\pars{-1} – \Re\mrm{f}\pars{\ic} – \Im\mrm{f}\pars{\ic} +
\left\{\begin{array}{rcl}
\ds{\mrm{f}\pars{z}} & \ds{\equiv} &
\ds{{1 \over 2}\sum_{n = 1}^{\infty}z^{n}\ln^{2}\pars{1 + {1 \over n}}}
\\[5mm]
\ds{\mrm{f}\pars{-1}} & \ds{\approx} & \ds{-0.1843}
\\[2mm]
\ds{\Re\mrm{f}\pars{\ic}} & \ds{\approx} & \ds{-0.0649}
\\[2mm]
\ds{\Im\mrm{f}\pars{\ic}} & \ds{\approx} & \ds{\phantom{-}0.2099}
\\[2mm]
\ds{{1 \over 2}\,\ln^{2}\pars{2}} & \ds{\approx} &\ds{\phantom{-}0.2402}
\end{array}\right.
\end{align}

We can even go further and write $\ds{\mrm{f}\pars{z}}$ as
\begin{align}
\mrm{f}\pars{z} & \equiv
{1 \over 2}\sum_{n = 1}^{\infty}z^{n}\ln^{2}\pars{1 + {1 \over n}} =
{1 \over 2}\sum_{n = 1}^{\infty}z^{n}\pars{\int_{0}^{1}{\dd x \over x + n}}
\pars{\int_{0}^{1}{\dd y \over y + n}}
\\[5mm] & =
{1 \over 2}\,\int_{0}^{1}\int_{0}^{1}\pars{%
\sum_{n = 1}^{\infty}{z^{n} \over n + x} –
\sum_{n = 1}^{\infty}{z^{n} \over n + y}}{\dd x\,\dd y \over y – x}
\\[5mm] & =
{1 \over 2}\,z\int_{0}^{1}\int_{0}^{1}
{\Phi\pars{z,1,x + 1} – \Phi\pars{z,1,y + 1} \over
y – x}\,\dd x\,\dd y
\end{align}

where $\ds{\Phi}$ is the
Lerch Transcendent Function. It still looks cumbersome !!!.

We can write
$$\log\left(1+\frac{1}{2n}\right)=\int_0^1 x^{2n-1}\frac{x-1}{\log(x)}\;dx\\ \log\left(1+\frac{1}{2n+1}\right)=\int_0^1 y^{2n}\frac{y-1}{\log(y)}\;dy$$
Then
$$S=\sum_{n=1}^\infty(-1)^n\log\left(1+\frac{1}{2n}\right)\log\left(1+\frac{1}{2n+1}\right)=\int_0^1\int_0^1 \sum_{n=1}^\infty x^{2n-1}y^{2n}\frac{x-1}{\log(x)}\frac{y-1}{\log(y)}\;dx\;dy$$
giving
$$S=-\int_0^1\int_0^1\frac{xy^2(x-1)(y-1)}{(1+x^2y^2)\log(x)\log(y)}\;dy\;dx \tag{1}$$
If we expand the integrand into a series
$$\frac{xy^2(x-1)(y-1)}{(1+x^2y^2)\log(x)\log(y)}=\left(1-\frac{1}{x}\right)\left(\sum_{k=1}^\infty \frac{(-1)^k(y-1)y^{2k}x^{2k}}{\log(x)\log(y)}\right)$$
and integrate over $y$ term by term we can get
$$S=\int_0^1 \frac{x(x-1)}{\log(x)}\left(\Phi^{(0,1,0)}(-x^2,0,2)-\Phi^{(0,1,0)}\left(-x^2,0,\frac{3}{2}\right)\right)\;dx$$
with the derivative of the Lerch Trancendent $\Phi(z,s,a)$ w.r.t $s$. They seem to check out numerically. I prefer equation (1) to that though.