# 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)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)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)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 $R ≥ \min \{ 1, 1\} = 1$.