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The radius of convergence of a power series is a radius of the largest disk where the series converges. It is always positive or $\infty$. Let us consider any power series
f(Z) = $\sum_{n=0}^{infty}$ cn (x-a)n.
The power series converges if |x-a|r.
Radius of Convergence of a power series calculator is a online tool for the convergence test. You just have to enter the power series to will get know whether the series is convergent or divergent.

You can see a calculator with default power series with its limits given below. The calculator will do convergence test by applying the limits and then go for divergent test if possible. You can see whether the series is convergent or divergent by clicking on "Calculate".

## Steps for Radius of Convergence Calculator

Step 1 :

Step 2 :

There are many ways to find radius of convergence. But here we are just considering ratio test. For the convergence test go for the ratio test and apply limits.If the series are not convergent go for the divergent test.

## Problems on Radius of Convergence Calculator

1. ### Find the radius of convergence of power series : $\sum_{n=1}^{\infty}$ $\frac{x^{2n}}{4^n}$

Step 1 :

The given power series is $\sum_{n=1}^{\infty}$ $\frac{x^{2n}}{4^n}$

Step 2 :

Applying limits to the power series we get
L = $\lim_{n \to \infty}\left|\frac{x^{2n}}{4^{n}}\right|^{\frac{1}{n}}$
= $\lim_{n \to \infty}\left|\frac{x^{2n}}{4^{n}}\right|$
= $lim_{n \to \infty}$ $\frac{x^2}{4}$
= $\frac{x^2}{4}$
The series converges if $\frac{x^2}{4}$ < 1.Hence $\frac{x^2}{4}$ < 1
x2 < 4 or |x| < 2
The series is not convergent.It converges only at the point |x|< 2. The interval of convergence is -2 < x < 2.
Lets go for divergence test. The given power series is
$\sum_{n=1}^{\infty}$ $\frac{2^{2n}}{-4^n}$  = $\sum_{n=1}^{\infty}$ $\frac{(2^{2)^{n}}}{-4^n}$
= $\sum_{n=1}^{\infty}$ $\frac{4^{n}}{(-1)^n 4^n}$
= $\sum_{n=1}^{\infty}$ (-1)n
Hence the series is divergent.

The power series $\sum_{n=1}^{\infty}$ $\frac{x^{2n}}{4^n}$ is divergent.

2. ### Find the radius of convergence for  the series $\sum_{n=1}^{\infty}$ $\frac{(x-4)^n}{n^n}$.

Step 1 :

The given power series is $\sum_{n=1}^{\infty}$ $\frac{(x-4)^n}{n^n}$

Step 2 :

Applying limits to the power series we get
L = $\lim_{n \to \infty}$ $\left|\frac{(x - 4)^{n}}{n^{n}}\right|^{\frac{1}{n}}$
= $\lim_{n \to \infty}$ $\left|\frac{(x - 4)}{n}\right|$
= |x-4| $lim_{n \to \infty}$ $\frac{1}{n}$
= 0
Since the limit L = 0 < 1. Hence the power series converges for every x. The radius of convergence is $\infty$ and the interval of convergence lies between - $\infty$ and + $\infty$.

The power series $\sum_{n=1}^{\infty}$ $\frac{(x-4)^n}{n^n}$ is convergent.