Question

How do you find the radius of convergence of a power series?

Answer 1

Step 1: The radius of convergence

In a power series, the radius of convergence is the largest disc at which it converges.

Generally, $\text{'}x\text{'}$ has to be a certain value in order for a power series to converge.

As an example, converges for $|x-a|<R$. Here $a,x$ are the numbers and $R$ is the radius of convergence.

Step 2: Radius of convergence by using ratio test.

$R$ is a radius of convergence that is positive and it converges if $|x-a|<R$ and diverges if $|x-a|>R$.

To calculate the radius of convergence, apply the ratio test to the absolute values.

Solve and find the number $\mathrm{x}$ for which $\left|\mathrm{x}\right|<\mathrm{R}$.
The interval of convergence is found by first finding the radius of convergence, and then checking the convergence of the series at the endpoints of the interval $(-\mathrm{R},\mathrm{R})$.

• Ratio test.

Ratio tests are used to determine whether the power series are converging or diverging.

• If the value of $x<1$ the series is absolutely convergent.
• If the value of $x>1$ the series is divergent.
• If the value of $x=1$ the series may be divergent, conditionally convergent, or absolutely convergent.

Example: $\sum _{n=1}^{\infty }\frac{x^n}{\sqrt{n}}$

Find the radius of converges of the series:

If $a_n=\frac{x^n}{\sqrt{n}}$
Then $\begin{array}{rcl}\frac{a_{n+1}}{a_n}& =& \left(\frac{x^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{x^n}\right)\end{array}$

Factor $x^n$ out of $x^{n+1}\sqrt{n}$:
$\Rightarrow a=\left(\frac{x^n\left(x\sqrt{n}\right)}{\sqrt{n+1}{x}^n}\right)$
Factor $x^n$ out of $\sqrt{n+1}{x}^n$:
$\begin{array}{rcl}& \Rightarrow & a=\left(\frac{x^n\left(x\sqrt{n}\right)}{x^n\sqrt{n+1}}\right)\\ & =& \left(\frac{x^n\left(x\sqrt{n}\right)}{x^n\sqrt{n+1}}\right)\left[\mathrm{Cancel}\mathrm{the}\mathrm{comon}\mathrm{factor}\right]\end{array}$

Apply the limit:

$\begin{array}{rcl}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|& =& \underset{n\to \infty }{\lim }\left|x\right|\frac{\sqrt{n}}{\sqrt{n+1}}\\ & =& \underset{n\to \infty }{\lim }\left|x\right|\frac{1}{\sqrt{1+\frac{1}{n}}}\\ & =& \left|x\right|\end{array}$

Step 3: Apply the ratio test.

By the ratio test, the given series converges when $\left|x\right|<1$ and the radius of convergence is $1$ and the interval is $(-1,1).$

Hence, the radius of convergence of a power series can be found by the ratio test.