Summing power series in closed form

First series. Differentiating $\sum x^n=\frac1{1-x}$: $\sum_{n\ge1}nx^{n-1}=\frac1{(1-x)^2}$. Multiplying by $x$:
$$\sum_{n\ge1}nx^n=\frac x{(1-x)^2}.$$

Second series. Differentiate again the series $\sum nx^n$, then multiply by $x$:
$$\sum_{n\ge1}n^2x^n=x\cdot\frac d{dx}\left(\frac x{(1-x)^2}\right)=x\cdot\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\frac{x(1+x)}{(1-x)^3}.$$

Numerical values.
- $x=\frac12$: $\sum\frac n{2^n}=\frac{1/2}{1/4}=2$.
- $x=\frac13$: $\sum\frac{n^2}{3^n}=\frac{\frac13\cdot\frac43}{\frac8{27}}=\frac{4}{9}\cdot\frac{27}8=\frac32$.

Remark. Probabilistic meaning: $\sum\frac n{2^n}=2$ is the expected number of tosses of a fair coin until the first head.