# Infinite series $\sum_{k=1}^\infty \frac{k}{2^k}$ and $\sum_{k=1}^\infty \frac{k^2}{2^k}$ [duplicate]

How to evaluate those infinite series? How are they called?

$$\sum_{k=1}^\infty \frac{k}{2^k} \quad \text{and} \quad \sum_{k=1}^\infty \frac{k^2}{2^k}$$

I'm really sorry for asking, but I can't figure out how to google such stuff when I even don't know the names/categories.

• "differentiated geometric series" Dec 17, 2014 at 14:34
• This may help you compute the sums. Dec 17, 2014 at 14:35
• @DavidMitra Thanks a lot, that's all I need to know ;-) Dec 17, 2014 at 14:37

$$\sum_{k=1}^{\infty}x^k=\frac{x}{1-x}\rightarrow\frac{d}{dx}\sum_{k=1}^{\infty}x^k=\sum_{k=1}^{\infty}kx^{k-1}=\frac{1}{(1-x)^2}\rightarrow \sum_{k=1}^{\infty}kx^{k}=\frac{x}{(1-x)^2}$$
Now let $$x=\frac{1}{2}$$, hence we have $$\sum_{k=1}^{\infty}k({\frac{1}{2}})^{k}=\frac{0.5}{(1-0.5)^2}=2$$
Now if we take the derivative, we will have $$\frac{d}{dx}\sum_{k=1}^{\infty}kx^{k}=\sum_{k=1}^{\infty}k^2x^{k-1}=\frac{d}{dx}\left(\frac{x}{(1-x)^2}\right)=\frac{1+x}{(1-x)^3}\rightarrow \sum_{k=1}^{\infty}k^2x^{k}=\frac{x(1+x)}{(1-x)^3}$$ Now if you let $$x=\frac{1}{2}$$, then you have: $$\sum_{k=1}^{\infty}k^2\left(\frac{1}{2}\right)^{k}=\frac{0.5(1+0.5)}{(1-0.5)^3}=\frac{1.5}{0.25}=6$$