Evaluating series $\sum_{n=0}^{\infty}\frac{n}{2^{n+1}\left ( n+1 \right )^2}$.

Evaluate the series:

$$\sum_{n=0}^{\infty}\frac{n}{2^{n+1}\left ( n+1 \right )^2}$$

Solution:



The dilogarithm function is defined as $ \displaystyle {\rm Li}_2(x)= -\int_0^x \frac{\ln(1-t)}{t}\,{\rm d}t$. . Expanding the $\ln (1-t) $ as a Taylor series and interchanging integral and sum we have what follows:

$$\begin{aligned}
{\rm Li}_2(x) &=-\int_{0}^{x}\frac{\ln(1-t)}{t}\,{\rm d}t \\
 &= \int_{0}^{x}\frac{1}{t}\sum_{n=1}^{\infty}\frac{t^n}{n}\,{\rm d}t\\
 &= \int_{0}^{x}\sum_{n=1}^{\infty}\frac{t^{n-1}}{n}\,{\rm d}t\\
 &=\sum_{n=1}^{\infty}\frac{1}{n}\int_{0}^{x}t^{n-1}\,{\rm d}t \\
 &=\sum_{n=1}^{\infty}\frac{x^n}{n^2}
\end{aligned}$$

Using the new formula we derived (expressing the dilog as a series) we get the following functional equation:

$$ \begin{equation} {\rm Li}_2(x)+{\rm Li}_2(1-x)=\frac{\pi^2}{6}-\ln x \ln (1-x), \; x \in (0, 1) \end{equation} $$

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 We define the function $\displaystyle f(x)=\sum_{k=0}^{\infty}\frac{kx^{k+1}}{\left ( k+1 \right )^2}$ and we differentiate it , thus

$$f'(x)=\sum_{k=0}^{\infty}\frac{kx^k}{k+1}=\sum_{k=0}^{\infty}x^k -\sum_{k=0}^{\infty}\frac{x^k}{k+1}=\frac{1}{1-x}+\frac{\ln (1-x)}{x}$$

Integrating back and plugging $x=1/2$ to get the wanted series we have:

$$f\left ( \frac{1}{2} \right )=\int_{0}^{1/2}\left [ \frac{1}{1-x}+\frac{\ln (1-x)}{x} \right ]\,{\rm d}x =\ln 2 -{\rm Li}_2\left ( \frac{1}{2} \right )$$

Using $(1)$ we have that :

$$ 2{\rm Li}_2 \left( \frac{1}{2} \right) =\frac{\pi^2}{6}-\ln^2 2 \Leftrightarrow {\rm Li}_2 \left( \frac{1}{2} \right)=\frac{\pi^2}{12}-\frac{\ln^2 2}{2}$$

Hence the initial series evaluates to:

$$\sum_{k=0}^{\infty}\frac{k}{2^{k+1}(k+1)^2}=\ln 2 +\frac{\ln^2 2}{2}-\frac{\pi^2}{12}$$