$\sum_{n=1}^{\infty}\frac{1}{n^22^n}$ by integration or differentiation

Solution 1:

Note that we have $\int_0^x t^{n-1}\,dt=\frac{x^n}{n}$. Then, we can write

$$\begin{align} \sum_{n=1}^\infty \frac{x^{2n}}{n^2}&=\sum_{n=1}^\infty \int_0^x t^{n-1}\,dt\int_0^x s^{n-1}\,ds\\\\ &=\int_0^x\int_0^x \frac{1}{1-st}\,ds\,dt\\\\ &=-\int_0^{x} \frac{\log(1-sx)}{s}\,ds\\\\ &=-\int_0^{x^2} \frac{\log(1-s)}{s}\,ds\\\\ &=\text{Li}_2(x^2) \end{align}$$

Evaluating at $x=1/\sqrt{2}$ yields

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n^2\,2^n}=\text{Li}_2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)}$$

And we are done!

To evaluate $\text{Li}_2(1/2)$, we exploit the relationship

$$\text{Li}_2(1-x)=-\text{Li}_2\left(1-\frac1x\right)-\frac12\log^2(x)$$

Letting $x=1/2$ yields

$$\text{Li}_2(1/2)=-\text{Li}_2\left(-1\right)-\frac12\log^2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)$$

where we used

$$\begin{align} \text{Li}_2(-1)&=-\int_0^{-1}\frac{\log(1-x)}{x}\,dx\\\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\\\\ &=\frac{\pi^2}{12} \end{align}$$

Solution 2:

Another way to calculate it: consider $$ f(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n^2}. $$ Differentiating w.r.t. $x$ gives $$ f'(x)=\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n}=\frac{1}{x}\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad xf'(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad (xf'(x))'=\sum_{n=1}^{+\infty}x^{n-1}=\frac{1}{1-x}. $$ Now integrating with $f(0)=0$ $$ xf'(x)=-\ln(1-x)\quad\Rightarrow\quad f(x)=-\int_0^x\frac{\ln(1-t)}{t}\,dt. $$ Motivation for termwise differentiation for power series is straightforward.