# Science:Infinite Series Module/Units/Unit 3/3.1 Power Series/3.1.05 Power Series Convergence Example

## Example

Determine the radius and interval of convergence of the infinite series

$\sum _{k=0}^{\infty }{\frac {k(x+3)^{k}}{4^{k+1}}}$ ## Complete Solution

### Step 1: Apply Ratio Test

The ratio test gives us:

{\begin{aligned}\lim _{k\rightarrow \infty }{\Bigg |}{\frac {(k+1)(x+3)^{k+1}}{4^{k+2}}}{\Bigg /}{\frac {k(x+3)^{k}}{4^{k+1}}}{\Bigg |}&=\lim _{k\rightarrow \infty }{\Bigg |}{\frac {(k+1)}{k}}\cdot {\frac {(x+3)^{k+1}}{(x+3)^{k}}}\cdot {\frac {4^{k+1}}{4^{k+2}}}{\Bigg |}\\&=\lim _{k\rightarrow \infty }{\Bigg |}{\frac {(k+1)}{k}}\cdot (x+3)\cdot {\frac {1}{4}}{\Bigg |}\\&={\Big |}{\frac {x+3}{4}}{\Big |}\end{aligned}} The ratio test tells us that the power series converges only when

${\Big |}{\frac {x+3}{4}}{\Big |}<1$ or $|x+3|<4$ . Therefore, the radius of convergence is 4.

### Step 2: Test End Points of Interval to Find Interval of Convergence

The inequality $|x+3|<4$ can be written as -7 < x < 1. By the ratio test, we know that the series converges on this interval, but we don't know what happens at the points x = -7 and x = 1.

At x = -7, we have the infinite series

$\sum _{k=0}^{\infty }{\frac {k(-4)^{k}}{4^{k+1}}}=\sum _{k=0}^{\infty }{\frac {k(-1)^{k}}{4}}$ This series diverges by the test for divergence.

At x = 1, we have the infinite series

$\sum _{k=0}^{\infty }{\frac {k4^{k}}{4^{k+1}}}=\sum _{k=0}^{\infty }{\frac {k}{4}}$ This series also diverges by the test for divergence.

Therefore, the interval of convergence is -7 < x < 1.

## Possible Challenges

### What Convergence Test Should Be Used?

For most problems, the ratio test can be used initially. If the ratio test yields an interval for the domain, we need to use other convergence tests to explore what the domain could be at the end points of the interval.