Radius of Convergence

Before learning about the radius of convergence, let’s recall what a power series is. In real analysis, power series is one of the most important types of series. For instance, we can employ them to describe transcendental functions like exponential functions, trigonometric functions, etc.

A power series is any series that can be written in the form:

\(\begin{array}{l}\sum_{n=0}^{\infty}c_{n}(x-a)^{n}\end{array} \)

Here, cn and a are the numbers. However, cn‘s are often referred to as the coefficients of the series. Also, we can say that the power series is the function of x.

Radius of Convergence of Series

Consider the power series:

\(\begin{array}{l}\sum_{n=0}^{\infty}c_{n}(x-a)^{n}\end{array} \)

The radius of convergence “R” is any number such that the power series will converge for |x – a| < R and diverge for |x – a| > R. The power series may not converge for |x – a| = R.

From this, we can define the interval of convergence as follows.

The interval of all x values, including the endpoints (if required) for which the power series converges, is called the interval of convergence of the series.

If the radius of convergence of the power series is R, then;

a – R < x < a + R (Power series converges)

x < a – R and x > a + R (Power series diverges)

Therefore, the radius of convergence of a power series will be half of the length of the interval of convergence. If “R” is the radius of convergence, then we can define the interval of convergence as (a − R, a + R).

Also, read:

How to Find the Radius of Convergence?

Using the Ratio test, we can find the radius of convergence of given power series as explained below.

\(\begin{array}{l}\sum_{n=0}^{\infty}c_{n}(x-a)^{n}\end{array} \)

Step 1: Let an = cn (x – a)n and an+1 = cn+1 (x – a)n+1

Step 2: Consider the limit for the absolute value of an+1/an as n → ∞.

i.e.,

\(\begin{array}{l}\displaystyle \lim_{n \to \infty}\left| \frac{a_{n+1}}{a_n}\right|\end{array} \)

Step 3: Simplify the ratio

\(\begin{array}{l}\frac{c_{n+1}.(x-a)^{n+1}}{c_n .(x-a)^n}\end{array} \)
, i.e., (1/cn)cn+1 (x – a) and apply the limit.

Step 4: Finally compute the result for R based on the scenarios given in the table below.

Lim value of the absolute ratio as n → ∞ R value
0 Infinite
The power series converges for all values of x.
∞ 0

The power series converges only at x = a.

N.|x – a|; N is any natural number R = 1/N

Try out:Radius of Convergence Calculator

The above process can be understood in a better way, using the solved examples given below.

Radius of Convergence Examples

Example 1:

Find the radius of convergence for the following.

\(\begin{array}{l}\sum_{n=1}^{\infty}\frac{(-1)^{n}.n(x+3)^n}{4^n}\end{array} \)

Solution:

Given power series:

\(\begin{array}{l}\sum_{n=1}^{\infty}\frac{(-1)^{n}.n(x+3)^n}{4^n}\end{array} \)

Let us find the radius of convergence using the ratio test.

Thus, an =

\(\begin{array}{l}\frac{(-1)^{n}.n(x+3)^n}{4^n}\end{array} \)
and an+1 =
\(\begin{array}{l}\frac{(-1)^{n+1}.(n+1)(x+3)^{n+1}}{4^{n+1}}\end{array} \)

Now, we have to take the limit as n → ∞ for the absolute ratio of an+1/an.

\(\begin{array}{l}\displaystyle \lim_{n \to \infty}\frac{a_{n+1}}{a_n}=\left|\frac{(-1)^{n+1}.(n+1)(x+3)^{n+1}}{4^{n+1}}\times \frac{4^n}{(-1)^n.n(x+3)^n}\right|\end{array} \)

This can be simplified as:

\(\begin{array}{l} =\displaystyle \lim_{n \to \infty}\left|\frac{(-1).(n+1)(x+3)}{4n}\right|\end{array} \)

Now, taking the term that contains x outside limit, we get;

\(\begin{array}{l}\displaystyle |x+3|\lim_{n \to \infty}\frac{(n+1)}{4n}\end{array} \)

= (¼) |x + 3|

This is of the form N.(x – a).

Therefore, the radius of convergence = R = 1/N = 1/(¼) = 4

Example 2:

Find the radius of convergence of e^x.

(or)

Show that the radius of convergence of ex is infinite

Solution:

We know that,

\(\begin{array}{l}e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+….\end{array} \)

Thus, the power series representation of ex is given by:

\(\begin{array}{l}\sum_{n=0}^{\infty}\frac{x^n}{n!}\end{array} \)

Using the ratio test, we can find the radius of convergence as:

\(\begin{array}{l}\displaystyle \lim_{n \to \infty}\left|\frac{x^{n+1}}{(n+1)!}\times \frac{n!}{x^n} \right|\end{array} \)

This can be simplified as:

\(\begin{array}{l} |x|\displaystyle \lim_{n \to \infty}\frac{1}{n+1}\end{array} \)

= 0

Therefore, the radius of convergence is infinite.

Practice Problems

  1. What is the radius of convergence for the series
    \(\begin{array}{l}h\sum_{n=0}^{\infty}2^{2n}x^n\end{array} \)
    ?
  2. Find the radius of convergence for the power series,
    \(\begin{array}{l}\sum_{n=0}^{\infty}n!(2x – 5)^n\end{array} \)
    .
  3. Determine the radius of convergence for the following series.
    \(\begin{array}{l}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{2n!}\end{array} \)

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