How to determine the sum of an infinite series that is neither arithmetic nor geometric?

Question

What is $$\sum_{k =1}^{\infty}\frac{k^2}{2^k}$$

Answer 1

We generally use differentiation techniques for this.

We have that for all $x$ with $|x|<1$ the following holds:

$$\sum_{k\geq 0}x^k=\frac{1}{1-x}$$

Now differentiate both sides of the equality above: the LHS term by term, and the RHS as you would in calculus. Equality still holds because of uniform convergence. This will net you something like $\sum kx^{k-1}$, which you can multiply by $x$ to obtain $\sum kx^k$.

You then repeat the process to obtain your desired series, and evaluate it at $x=\frac12$.

Answer 2

This is not the best way to find the value, but it shows that with some effort it can be evaluated:

\begin{align*} S= \sum_{k=1}^\infty \frac{k^2}{2^k} &= \frac{1}{2} + \frac 4 4 + \frac 9 8 + \frac{16}{16} + \frac{25}{32} + \frac{36}{64} + \frac{49}{128} + \cdots \\ \\ &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \cdots \\ \\ &\quad \quad +\frac{3}{4} + \frac{3}{8} + \frac{3}{16}+ \frac{3}{32} + \frac{3}{64} + \frac{3}{128}+ \cdots \\ \\ &\quad \quad \quad \quad+ \frac{5}{8} + \frac{5}{16}+ \frac{5}{32} + \frac{5}{64} + \frac{5}{128}+ \cdots \\ \\ & \quad \quad \quad \quad \quad \quad \vdots \\ \\ \\ &= 1\cdot 1 + 3 \cdot \frac{1}{2} + 5 \cdot \frac{1}{4} + 7 \cdot \frac{1}{8} + \cdots \\ \\ &= \sum_{n=0}^\infty \frac{2n+1}{2^n} = 2\cdot \underbrace{\sum_{n=0}^\infty \frac{n}{2^n}}_{=A}+ \underbrace{\sum_{n=1}^\infty \frac{1}{2^n}}_{=2} =S \end{align*}

So, $S=2A +2$.

\begin{align*} A&= \frac{1}{2}+ \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{32} + \cdots \\ \\ &= \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \cdots \\ \\ & \quad \quad +\frac{1}{4}+ \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \cdots \\ \\ & \quad \quad \quad \quad +\frac{1}{8} + \frac{1}{16} + \frac{1}{32} \cdots \\ \\ & \quad \quad \quad \quad \quad \quad + \cdots \\ \\ &=1 + \frac{1}{2} + \frac{1}{4} + \cdots = 2=A \end{align*}

Therefore $S=2A+2 = 6$

Answer 3

hint: use induction to show that for your finite sum is hold $$\sum_{k=1}^m\frac{k^2}{2^k}=-4\, \left( 1/2 \right) ^{m+1} \left( m+1 \right) -6\, \left( 1/2 \right) ^{m+1}-2\, \left( 1/2 \right) ^{m+1} \left( m+1 \right) ^{2}+ 6 $$ note that for $m\to\infty$ we get $6$