What if the limit L evaluates to 0?

If L = 0 for all x, the inequality 0 < 1 is trivially true for all real numbers. The series converges everywhere, and the radius of convergence is R = \infty.

What if the limit L evaluates to infinity?

If L = \infty for any x eq a, the series only converges at its center x = a. The radius of convergence is R = 0.

When should the Root Test be used instead of the Ratio Test?

The Root Test is strictly preferred when the general term a_n involves n-th powers across all components, making the limit \lim_{n o \infty} |a_n|^{1/n} algebraically simpler to evaluate than the ratio.

How to find the radius of convergence of a power series

Q: What if the limit L evaluates to infinity?

If L = \infty for any x eq a, the series only converges at its center x = a. The radius of convergence is R = 0.

Q: When should the Root Test be used instead of the Ratio Test?

The Root Test is strictly preferred when the general term a_n involves n-th powers across all components, making the limit \lim_{n o \infty} |a_n|^{1/n} algebraically simpler to evaluate than the ratio.

The radius of convergence of a power series $\sum c_n (x-a)^n$ is found by applying the Ratio Test or the Root Test to the terms of the series. This method applies to any power series to determine the distance $R$ from the center $a$ within which the series converges absolutely.

The setup

Identify the general term of the series, $a_n = c_n (x-a)^n$ . Note the center of the series, $a$ , and ensure the series is written in standard power series form.

The steps

Form the ratio of consecutive terms and take the absolute value: $|a_{n+1} / a_n|$ . 2. Evaluate the limit as $n o \infty$ of this ratio to find $L$ . 3. By the Ratio Test, the series converges absolutely when $L < 1$ . Set up the inequality $L < 1$ . 4. Isolate the absolute value expression containing $x$ , resulting in the form $|x-a| < R$ . The constant $R$ is the radius of convergence.

Checking the result

Verify that $R$ is a non-negative real number or $\infty$ . If $L = 0$ for all $x$ , then $R = \infty$ . If $L = \infty$ for all $x eq a$ , then $R = 0$ . Check that algebraic simplification of factorials and exponents was done correctly.

Common errors

Forgetting the absolute value signs in the limit, which can lead to incorrect domain assumptions. Confusing the radius of convergence $R$ with the interval of convergence (finding the interval requires testing the endpoints $x = a-R$ and $x = a+R$ separately). Dropping the exponent $n$ incorrectly during algebraic simplification.

Worked example

Find the radius of convergence of $\sum_{n=1}^{\infty} \frac{(-1)^n (x-3)^n}{n 2^n}$ .

Let $a_n = \frac{(-1)^n (x-3)^n}{n 2^n}$ . Compute the limit $L = \lim_{n o \infty} \left| \frac{a_{n+1}}{a_n} \right|$ . Substitution gives $L = \lim_{n o \infty} \left| \frac{(-1)^{n+1} (x-3)^{n+1}}{(n+1) 2^{n+1}} \cdot \frac{n 2^n}{(-1)^n (x-3)^n} \right|$ . Simplify the absolute value: $L = \lim_{n o \infty} \left| \frac{(x-3) \cdot n}{(n+1) \cdot 2} \right|$ . Factor out the terms independent of $n$ : $L = \frac{|x-3|}{2} \lim_{n o \infty} \frac{n}{n+1}$ . Evaluate the limit: $L = \frac{|x-3|}{2} (1) = \frac{|x-3|}{2}$ . Set $L < 1$ for convergence: $\frac{|x-3|}{2} < 1$ . Multiply by 2: $|x-3| < 2$ . The radius of convergence is $R = 2$ .

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References: Calculus: Early Transcendentals by James Stewart · OpenStax Calculus Volume 2