AcademyPower Series

Academy

Radius of Convergence

Level 1 - Math I (Physics) topic page in Power Series.

Principle

The radius of convergence tells how far from the centre a power series can be trusted before it stops converging. For a series centred at \(c\), the main convergence region has the form \(|x-c|\lt R\).

Radius condition
\[|x-c|\lt R\]

The radius is a distance in the input variable. It does not decide the endpoints \(x=c-R\) and \(x=c+R\); those must be tested separately.

Notation

\(R\)
radius of convergence
\(c\)
centre of the power series
\(x\)
input variable
\(u=x-c\)
shifted variable
\(t_n\)
nth term of the power series

The value \(R\) may be finite, zero, or infinite. In Level 1 examples it is usually found using the ratio test.

Method

Step 1: Write the nth term

For

General term
\[\sum_{n=0}^{\infty}a_n(x-c)^n\]

the nth term is \(t_n=a_n(x-c)^n\).

Step 2: Apply the ratio test

Use the absolute ratio of consecutive terms:

nth term
\[t_n=a_n(x-c)^n\]
Next term
\[t_{n+1}=a_{n+1}(x-c)^{n+1}\]
Absolute ratio
\[\left|\frac{t_{n+1}}{t_n}\right|=\left|\frac{a_{n+1}}{a_n}\right||x-c|\]
Convergence requirement
\[\lim_{n\to\infty}\left|\frac{t_{n+1}}{t_n}\right|\lt1\]

Step 3: Solve for the distance from the centre

The inequality from the ratio test usually becomes \(|x-c|\lt R\). The number on the right is the radius of convergence.

Rules

Ratio-test condition
\[\lim_{n\to\infty}\left|\frac{t_{n+1}}{t_n}\right|\lt1\]
Finite radius form
\[c-R\lt x\lt c+R\]
Infinite radius
\[R=\infty\]

If the ratio limit is \(L|x-c|\), then convergence requires \(L|x-c|\lt1\), so \(R=1/L\) when \(L>0\).

Examples

Question
Find the radius of convergence of
\[\sum_{n=0}^{\infty}x^n\]
Answer
The ratio is
\[|x^{n+1}/x^n|=|x|\]
Convergence requires
\[|x|\lt1\]
so the radius is
\[R=1\]

Checks

  • The radius is a non-negative distance, not an interval.
  • A ratio test result of \(|x-c|\lt R\) says nothing final about endpoints.
  • If \(R=\infty\), the series converges for every real \(x\).
  • If \(R=0\), the series converges only possibly at its centre.