The Radius of Convergence of a Power Series Examples 3

Recall from The Radius of Convergence of a Power Series page that we can calculate the radius of convergence of a power series $\sum_{n=0}^{\infty} a_n(x - c)^n$ 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 $\sum_{n=0}^{\infty} \frac{3^n}{5^n \sqrt{n}} x^n$.

We have that:

(1)

\begin{align} \quad \lim_{n \to \infty} \biggr \vert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \biggr \rvert \frac{3^{n+1}}{5^{n+1} \sqrt{n+1}} \frac{5^n \sqrt{n}}{3^n} \biggr \rvert = \lim_{n \to \infty} \frac{3}{5} \frac{\sqrt{n}}{\sqrt{n+1}} = \frac{3}{5} \end{align}

Therefore the radius of convergence is $\frac{5}{3}$.

Example 2

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

We have that:

(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{2^{n+1} (n+1)! (n+1)^2 (3 + \sqrt{n+1})}{2 \cdot 3^{n+3} (n + 2)^2} \frac{2 \cdot 3^{n+2} (n + 1)^2}{2^n n! n^2 (3 + \sqrt{n})}\biggr \rvert = \lim_{n \to \infty} \frac{2}{3} \frac{(n+1)^5(3 + \sqrt{n+1})}{(n+2)^2n^2(3 + \sqrt{n})} = \infty \end{align}

The limit above equals $\infty$ because the degree in the numerator is greater than the degree in the denominator. Therefore the radius of convergence is $0$.