Evaluating Sums with Power Series

# Evaluating Sums with Power Series

We will now look at an application of power series - namely in finding the sums of relatively complicated series.

## Example 2

Find the sum of the series $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$ by finding the sum of the power series $\sum_{n=1}^{\infty} n^2x^n$.

Let's first look at the power series $f(x) = \frac{1}{1 - x} = \sum_{n=0}^{\infty} x_n$ for $-1 < x < 1$, which when differentiated produces the series $f'(x) \frac{1}{(1 - x)^2} = \sum_{n=1}^{\infty} nx^{n-1}$.

We will take this series and multiply it by $x$ so that:

(1)
\begin{align} \frac{x}{(1 - x)^2} = \sum_{n=1}^{\infty} nx^{n-1} \\ \frac{x}{(1 - x)^2} = \sum_{n=1}^{\infty} nx^n \end{align}

Now we will differentiate both sides to get the following series:

(2)
\begin{align} \frac{d}{dx} \frac{x}{(1 - x)^2} = \sum_{n=1}^{\infty} n^2x^{n-1} \\ \frac{1 + x}{(1 - x)^3} = \sum_{n=1}^{\infty} n^2x^{n-1} \end{align}

Once again, we will multiply both sides by $x$ to get:

(3)
\begin{align} \frac{x(1+x)}{(1-x)^3} = \sum_{n=1}^{\infty} n^2x^n \end{align}

Now let $x = \frac{1}{2}$, and so:

(4)
\begin{align} \frac{\frac{1}{2} \left ( 1 + \frac{1}{2} \right )}{\left ( 1 - \frac{1}{2} \right )^3} = \sum_{n=1}^{\infty} n^2 \left ( \frac{1}{2} \right )^n = \sum_{n=1}^{\infty} \frac{n^2}{2^n} = 6 \end{align}