Computing Sums with Power Series

Recall from the Determining a Function Representing a Power Series page that sometimes we would like to take a given series and find a function $f$ for which this series converges to. If we are able to find such a function - then evaluating the power series for $x$ in which the power series converges is just a matter of evaluating the function $f$ at that value of $x$.

We will now look at some examples of using power series in order to evaluate exact sums of trickier series (recall that we have only been able to compute exact sums of geometric series, telescoping series, and estimate sums for alternating series thus far).

Example 1

Using power series, find the sum of $\sum_{n=1}^{\infty} \frac{n}{3^n}$.

Consider the power series $\sum_{n=0}^{\infty} nx^n$. If we evaluate this power series at $x = \frac{1}{3}$, then we obtain the series given in the problem (apart from a different start index).

Suppose that $f(x) = \sum_{n=0}^{\infty} nx^n$. Then:

(1)

\begin{align} \quad f(x) = \sum_{n=0}^{\infty} nx^n \\ \quad \frac{f(x)}{x} = \sum_{n=0}^{\infty} nx^{n-1} \\ \quad \int \frac{f(x)}{x} \: dx = C + \sum_{n=0}^{\infty} x^n \\ \quad \int \frac{f(x)}{x} \: dx = C + \frac{1}{1 - x} \\ \quad \frac{d}{dx} \int \frac{f(x)}{x} \: dx =-\frac{1}{(1 - x)^2} \\ \quad \frac{f(x)}{x} = \frac{1}{(1 - x)^2} \\ \quad f(x) = \frac{x}{(1 - x)^2} \end{align}

Therefore $f(x) = \frac{x}{(1 - x)^2} = \sum_{n=0}^{\infty} nx^n$. This series converges for $\mid x \mid < 1$, in other words, $- 1 < x < 1$. Plugging in $x = \frac{1}{3}$ and we have that:

(2)

\begin{align} \quad f\left ( \frac{1}{3} \right ) = \frac{\frac{1}{3}}{\left ( \frac{2}{3} \right )^2} = \sum_{n=0}^{\infty} \frac{n}{3^n} \\ \quad \sum_{n=0}^{\infty} \frac{n}{3^n} = \sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4} \end{align}

Example 2

Using power series, find the sum of $\sum_{n=0}^{\infty} \frac{(n+1)^2}{\pi^n}$.

Consider the power series $\sum_{n=0}^{\infty} (n + 1)^2 x^n$. Let $f(x) = \sum_{n=0}^{\infty} (n + 1)^2 x^n$. We want to solve for $f(x)$:

(3)

\begin{align} \quad f(x) = \sum_{n=0}^{\infty} (n + 1)^2 x^n \\ \quad \int f(x) \: dx = C + \sum_{n=0}^{\infty} (n+1)x^{n+1} \\ \quad \frac{\int f(x) \: dx}{x} = \frac{C}{x} + \sum_{n=0}^{\infty} (n+1) x^n \\ \quad \int \frac{\int f(x) \: dx}{x} \: dx = D + \int \frac{C}{x} \: dx + \sum_{n=0}^{\infty} x^{n+1} \\ \quad \int \frac{\int f(x) \: dx}{x} \: dx = D + \int \frac{C}{x} \: dx + x \sum_{n=0}^{\infty} x^n \\ \quad \int \frac{\int f(x) \: dx}{x} \: dx = D + \int \frac{C}{x} \: dx + \frac{x}{1 - x} \\ \quad \frac{\int f(x) \: dx}{x} = \frac{C}{x} + \frac{(1 - x)(1) - x(-1)}{(1 - x)^2} \\ \quad \frac{\int f(x) \: dx}{x} = \frac{C}{x} + \frac{1}{(1 - x)^2} \\ \quad \int f(x) \: dx = C + \frac{x}{(1 - x)^2} \\ \quad f(x) = \frac{(1 - x)^2(1) - x(2(1 - x)(-1))}{(1 - x)^4} \\ \quad f(x) = \frac{(1 - x)^2 + 2x(1 - x)}{(1 - x)^4} \\ \quad f(x) = \frac{(1 - x) + 2x}{(1 - x)^3} \end{align}

This series converges for $\mid x \mid < 1$, that is, for $- 1 < x < 1$. Plugging in $x = \frac{1}{\pi}$ (which is a value of $x$ contained in the interval of convergence) and we get that:

(4)

\begin{align} \quad f \left ( \frac{1}{\pi} \right ) = \sum_{n=0}^{\infty} \frac{(n + 1)^2}{\pi^n} = \frac{\left (1 - \frac{1}{\pi} \right ) + \frac{2}{\pi}}{\left (1 - \frac{1}{\pi} \right )^3} \end{align}