# May 28, 2021 - 03:04

#### An infinite series problem.

By Vamshi Jandhyala in mathematics

May 28, 2021

## Problem

— 級数bot (@infseriesbot) May 28, 2021

## Solution

We have

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

Multiplying by $x^2$ on both sides we get

$$
\begin{aligned}
\frac{x^2}{(1-x)^2} &= \sum_{k=0}^\infty (k+1)x^{k+2} \\

\end{aligned}
$$

Substituting $x= \frac{1}{2}$ in the above identity, we get

$$ \begin{aligned} \sum_{k=0}^\infty \frac{k}{2^{k+1}} = 1 \end{aligned} $$

Differentiating and mutliplying the equation below by $x^2$

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

we get

$$ \sum_{k=0}^\infty (k+1)^2 x^{k+2} = \frac{x^2(x+1)}{(1-x)^3} $$

Substituting $x= \frac{1}{2}$ in the above identity, we get

$$ \begin{aligned} \sum_{k=0}^\infty \frac{k^2}{2^{k+1}} = 3 \end{aligned} $$

The last two identities can be obtained using the same approach as above.

All the identities below are a generalization of the above identities

— 級数bot (@infseriesbot) June 7, 2021