Algebraic Operations of Power Series Examples 1

# Algebraic Operations of Power Series Examples 1

Recall from the Algebraic Operations of Power Series page that:

• If the power series $\sum_{n=0}^{\infty} a_n x^n$ has radius of convergence $R_1$ and if $k \in \mathbb{R}$ is some constant, then the power series $\sum_{n=0}^{\infty} (ka_n)x^n$ also has radius of convergence $R_1$ and $\sum_{n=0}^{\infty} (ka_n) x^n = k \sum_{n=0}^{\infty} a_n x^n$.
• If the power series $\sum_{n=0}^{\infty} a_n x^n$ has radius of convergence $R_1$ and the power series $\sum_{n=0}^{\infty} b_n x^n$ has radius of convergence $R_2$, then the power series $\sum_{n=0}^{\infty} (a_n + b_n) x^n$ has radius of convergence $R ≥ \min \{ R_1, R_2 \}$ and whenever both $\sum_{n=0}^{\infty} a_n x^n$ and $\sum_{n=0}^{\infty} b_n x^n$ converge, we have that $\sum_{n=0}^{\infty} (a_n + b_n)x^n = \sum_{n=0}^{\infty} a_nx^n + \sum_{n=0}^{\infty} b_nx^n$.

We will now look at some examples regarding these algebraic operations of power series.

## Example 1

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

Using the ratio test and 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} \frac{2^{n+1} (n+1)!}{(n+1)^2} \frac{n^2}{2^n n!} = \lim_{n \to \infty} \frac{2(n+1) n^2}{(n + 1)^2} = \lim_{n \to \infty} \frac{2n^2}{n + 1} = \infty \end{align}

Therefore the radius of convergence is $0$.

Furthermore, $\sum_{n=0}^{\infty} \frac{4 \cdot 2^n n!}{n^2} (x - 1)^2 = 4 \sum_{n=0}^{\infty} \frac{2^n n!}{n^2} (x - 1)^2$, so the radius of convergence of $\sum_{n=0}^{\infty} \frac{4 \cdot 2^n n!}{n^2} (x - 1)^2$ is $4(0) = 0$.

## Example 2

Determine the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{2\ln (n+1)}{n + 3} (x + 4)^n$ and the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n^2 - 2n + 1}{n + 1} (x + 4)^n$. Give a lower bound for the radius of convergence of the series $\sum_{n=0}^{\infty} \left ( \frac{2\ln (n + 1)}{n + 3} + \frac{n^2 - 2n + 1}{n + 1} \right ) (x + 4)^n$.

Applying the ratio test to the first series gives us:

(2)
\begin{align} \quad \lim_{n \to \infty} \biggr \rvert \frac{a_{n+1}}{a_n} \biggr \rvert = \lim_{n \to \infty} \frac{2\ln (n+2)}{n + 4} \frac{n+3}{2 \ln (n + 1)} = 1 \end{align}

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

Applying the ratio test to the second series gives us:

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

Therefore the radius of convergence of the power series $\sum_{n=0}^{\infty} \frac{n^2 - 2n + 1}{n + 1} (x + 4)^n$ is $1$ as well.

The radius of convergence of the power series $\sum_{n=0}^{\infty} \left ( \frac{2\ln (n + 1)}{n + 3} + \frac{n^2 - 2n + 1}{n + 1} \right ) (x + 4)^n$ will be at least as largest as the smallest radius of convergence of each individual term in the sum, that is, if $R$ is the radius of convergence of this series then $R ≥ \min \{ 1, 1\} = 1$.