The Radius of Convergence of a Power Series Examples 1

Recall from The Radius of Convergence of a Power Series page that we can calculate the radius of convergence of a power series using the ratio test, that is if $\lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = L$, then the radius of convergence is $R = \frac{1}{L}$. If $L = 0$ then the radius of convergence $R = \infty$ and if $L = \infty$ then the radius of convergence $R = 0$.

We will now look at some more examples of determining the radius of convergence of a given power series.

Example 1

Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

We first notice that $a_n = \frac{1}{n!}$ in our power series, and applying the rule above we have that

(1)

\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{n!}{(n+1)!} = \lim_{n \to \infty} \frac{1}{n+1} = 0 \end{align}

Therefore the radius of convergence of this power series is $R = \infty$.

Example 2

Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{(3x + 4)^n}{(n^3 + 2)3^n}$.

Let's first rewrite this power series as $\sum_{n=0}^{\infty} \frac{(3x + 4)^n}{(n^3 + 2)3^n} = \sum_{n=0}^{\infty} \frac{1}{n^3 + 2} \left ( \frac{3x + 4}{3} \right )^n = \sum_{n=0}^{\infty} \frac{1}{n^3 + 2} \left ( x + \frac{4}{3} \right )^n$.

Now we note that $a_n = \frac{1}{n^3 + 2}$, and using the ratio test:

(2)

\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_{n}} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{\frac{1}{(n+1)^3 + 2}}{\frac{1}{n^3 + 2}} \biggr \rvert = \lim_{n \to \infty} \frac{n^3 + 2}{(n+1)^3 + 2} = 1 \end{align}

So the radius of convergence is $R = \frac{1}{L} = 1$.

Example 3

Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n!(5x + 3)^n}{(n+1)^2 + 4n}$.

Let's first rewrite this series as $\sum_{n=0}^{\infty} \frac{n!(5x + 3)^n}{(n+1)^2 + 4n} = \sum_{n=0}^{\infty} \frac{n!}{(n+1)^2 + 4n} (5x + 3)^n = \sum_{n=0}^{\infty}\frac{n!}{(n+1)^2 + 4n} \left (5 \left (x + \frac{3}{5} \right ) \right)^n = \sum_{n=0}^{\infty}\frac{5^nn!}{(n+1)^2 + 4n} \left (x + \frac{3}{5} \right)^n$.

Now we can see that $a_n = \frac{5^nn!}{(n+1)^2 + 4n}$. Using the ratio test to find the radius of convergence we have:

(3)

\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{\frac{5^{n+1}(n+1)!}{(n+2)^2 + 4(n+1)}}{\frac{5^nn!}{(n+1)^2 + 4n}} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{5^{n+1}(n+1)!\left [ (n+1)^2 +4n \right ]}{\left [ (n+2)^2 + 4(n+1)\right] 5^n n!} \biggr \rvert = \lim_{n \to \infty} \frac{5(n+1)[(n+1)^2 + 4n]}{[(n+2)^2 + 4(n+1)]} = \infty \end{align}

Therefore the radius of convergence $R = 0$.