Power Series: Radius of Convergence, Solved Examples

A series of the form a₀+a₁x+a₂x²+… = ∑ a_nxⁿ is called a power series. The radius of convergence of a power series is a very powerful tool in many areas of Mathematics. In this article, we will study the convergence of a power series.

Definition of Power Series

A series of the form

a₀+a₁(x-a)+a₂(x-a)²+… = $\displaystyle \sum_{n=0}^\infty$ a_n(x-a)ⁿ,

where a₁, a₂, a₃, … are real numbers, is called a power series about the point x=a. This is the general form of a power series.

The general form of power series can be reduced to the following form:

a₀+a₁y+a₂y²+… = $\displaystyle \sum_{n=0}^\infty$ a_nyⁿ

by replacing y=x-a. This is a power series about the origin x=0.

Examples of Power Series

The following are examples of power series.

1. 1+x+x²+x³+… = $\displaystyle \sum_{n=0}^\infty$ xⁿ
2. $1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots$ = $\displaystyle \sum_{n=0}^\infty \dfrac{x^n}{n!}$.

Note that the first series is equal to $\dfrac{1}{1-x}$ provided that |x|<1, and the second series converges for all values of x in ℝ.

Radius of Convergence

Definition: A power series a₀+a₁x+a₂x²+… = ∑ a_nxⁿ is said to have R as radius of convergence if the power series converges absolutely for |x|<R and diverges for |x|>R.

Note that

• If R=∞, then the series is everywhere convergent.
• If R=0, then the series is nowhere convergent.

Interval of Convergence

If R is the radius of convergence of a power series, then the open interval (-R, R) is called the interval of convergence of the series.

For example, the geometric series 1+x+x²+x³+… = $\displaystyle \sum_{n=0}^\infty$ xⁿ converges absolute when |x|<1 and diverges when |x|>1. Thus, by definition, its radius of convergence is equal to 1 and the interval of convergence = (-1, 1).

To find the radius of convergence of a power series, we will use the following two tests: ratio test and root test.

Ratio Test

By ratio test, the radius of convergence R of a power series a₀+a₁x+a₂x²+… = ∑ a_nxⁿ is given by

$\dfrac{1}{R}$ = lim_n→∞ $\left| \dfrac{a_{n+1}}{a_n}\right|$.

Cauchy-Hadamard Test

Statement: If R is the radius of convergence of a power series ∑ a_nxⁿ, then we have:

$\dfrac{1}{R}$ = lim_n→∞ $\left| a_n\right|^{1/n}$.

Solved Examples

Q1: Find the radius of convergence of $x+\dfrac{2^2x^2}{2!}+\dfrac{3^3x^3}{3!}+\cdots$.

Answer:

The given series = $\displaystyle \sum_{n=0}^\infty$ a_nxⁿ.

Here a₀ = 0, a_n = $\dfrac{n^n}{n!}$ for n = 1, 2, 3, …

Now, if R is the radius of convergence of the given power series then by ratio test we have

$\dfrac{1}{R}$ = lim_n→∞ $\left| \dfrac{a_{n+1}}{a_n}\right|$

= lim_n→∞ $\left| \dfrac{(n+1^{n+1}}{(n+1)!} \times \dfrac{n^n}{n!}\right|$

= lim_n→∞ $\left(\dfrac{n+1}{n} \right)^n$

= lim_n→∞ $\left(1+\dfrac{1}{n} \right)^n$

= e.

Therefore, the radius of convergence of the power series is equal to 1/e.

Q2: Find the radius of convergence of $\displaystyle \sum_{n=0}^\infty 2^{2n}x^{n^2}$.

Answer:

Assume that n²=N ⇒ n=√N.

So the given series = $\displaystyle \sum_{N=0}^\infty 2^{2\sqrt{N}}x^{N}$.

Thus, a_N = $2^{2\sqrt{N}}$.

Now, the radius of convergence R by Cauchy-Hadamard test is given by

$\dfrac{1}{R}$ = lim_N→∞ $\left| a_N\right|^{\frac{1}{N}}$

= lim_N→∞ $\left| 2^{2\sqrt{N}}\right|^{\frac{1}{N}}$

= lim_N→∞ $\left| 2^{\frac{2}{\sqrt{N}}} \right|$

= 2⁰

= 1.

So the radius of convergence of the power series is equal to 1.

Also Study:

• Convergence of a Series
• Comparison Test for Series [Limit Form]
• Ratio Test for Series
• Root Test
• Alternating Series and Leibnitz’s Test
• Absolute and Conditional Convergence

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