Finding Radius Of Convergence Power Series

Finding the Radius of Convergence of a Power Series

A power series is an infinite sum of the form (\displaystyle \sum_{n=0}^{\infty} a_n (x-c)^n), where (a_n) are coefficients, (c) is the center, and (x) is the variable. Determining the radius of convergence tells us for which values of (x) the series converges to a finite sum, and it is a fundamental step in analyzing functions represented by series, solving differential equations, and approximating functions with Taylor or Maclaurin expansions. In this guide we will explore the concept of radius of convergence, walk through the most reliable methods to compute it, illustrate each technique with worked examples, and highlight common pitfalls to avoid.

Introduction to Power Series and Convergence

Before diving into the calculations, it helps to recall why convergence matters. A power series may behave like a polynomial for some (x) values and diverge (blow up to infinity or oscillate) for others. The set of all (x) where the series converges is called the interval of convergence. When the series is centered at (c), this interval is symmetric around (c) and can be expressed as ((c-R,,c+R)), where (R\ge 0) is the radius of convergence. If (R=0), the series converges only at (x=c); if (R=\infty), it converges for every real (or complex) (x).

The radius of convergence is not just a theoretical curiosity; it determines the domain of analyticity for the function represented by the series and guides numerical approximations. Consequently, mastering the techniques to find (R) is essential for anyone studying calculus, analysis, or applied mathematics.

Definition of Radius of Convergence

For a power series (\displaystyle \sum_{n=0}^{\infty} a_n (x-c)^n), the radius of convergence (R) is defined by

[ \frac{1}{R}= \limsup_{n\to\infty} \sqrt[n]{|a_n|}, ]

with the conventions that (R=0) when the limit superior is (\infty) and (R=\infty) when it is (0). This formula, known as the Cauchy–Hadamard theorem, provides a direct way to compute (R) from the coefficients alone. In practice, we often use simpler tests—the ratio test or the root test—because they avoid dealing with (\limsup) explicitly when the limit exists.

Methods to Find the Radius of Convergence

1. Ratio Test (D’Alembert’s Test)

If the limit

[ L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| ]

exists (or can be evaluated as (\infty) or (0)), then the radius of convergence is

[ R = \frac{1}{L}. ]

When (L=0), the series converges for all (x) ((R=\infty)). When (L=\infty), the series converges only at (x=c) ((R=0)).

Why it works: The ratio test examines the behavior of successive terms; if the ratio of magnitudes settles to a constant (L), the series behaves like a geometric series with ratio (L|x-c|). Convergence requires (L|x-c|<1), leading to (|x-c|<1/L).

2. Root Test (Cauchy’s Test)

Define

[ L = \lim_{n\to\infty} \sqrt[n]{|a_n|}. ]

If this limit exists, then

[ R = \frac{1}{L}. ]

The root test is especially useful when coefficients involve factorials or powers that simplify under an (n)‑th root.

3. Cauchy–Hadamard Formula (General Case)

When the ratio or root limits do not exist, we fall back on

[ \frac{1}{R}= \limsup_{n\to\infty} \sqrt[n]{|a_n|}. ]

The (\limsup) always exists (possibly as (\infty)), guaranteeing a value for (R). In most textbook problems, the limit exists, so the ratio or root test suffices.

Step‑by‑Step Procedure

1. Write the series in standard form (\displaystyle \sum a_n (x-c)^n). Identify (a_n) and the center (c).
2. Choose a test:
   • If (a_n) contains factorials, exponentials, or simple polynomials, try the ratio test. - If (a_n) involves (n)-th powers (e.g., (a_n = (b_n)^n)), the root test may be simpler.
   • If the limit is elusive, compute the (\limsup) directly.
3. Compute the limit (L) (ratio or root). 4. Determine (R) using (R = 1/L) (with the special cases (L=0\Rightarrow R=\infty) and (L=\infty\Rightarrow R=0)).
4. State the interval of convergence as ((c-R,,c+R)) and, if needed, test the endpoints separately (the radius test does not guarantee convergence at (x=c\pm R)).

Worked Examples

Example 1: Simple Polynomial Coefficients

Find the radius of convergence for

[ \sum_{n=0}^{\infty} \frac{(x-2)^n}{n+1}. ]

Solution

• Here (a_n = \frac{1}{n+1}), center (c=2).
• Apply the ratio test:

[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{1/(n+2)}{1/(n+1)} = \frac{n+1}{n+2} \xrightarrow[n\to\infty]{} 1. ]

Thus (L=1) and

[ R = \frac{1}{L}=1. ]

The series converges for (|x-2|<1), i.e., (x\in(1,3)). Endpoint testing (optional) shows convergence at (x=3) (alternating harmonic series) and divergence at (x=1) (harmonic series).

Example 2: Factorial Coefficients

Determine (R) for

[\sum_{n=0}^{\infty} \frac{n!}{(2n)!}, x^n. ]

Solution

• (a_n = \frac{n!}{(2n)!}), center (c=0).
• Ratio test:

[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{(2n+2)!}\cdot\frac{(2n)!}{n!} = \frac{n+1}{(2n+2)(2n+1)}. ]

As (n\to\infty), the numerator grows linearly while the denominator grows quadratically, so the limit is (0). Hence (L=0) and

Hence (L=0) and therefore

[ R=\frac{1}{L}= \infty . ]

The series (\displaystyle \sum_{n=0}^{\infty}\frac{n!}{(2n)!}x^n) converges for every real (or complex) (x); its interval of convergence is ((-\infty,\infty)).

Example 3: Root Test with Powers

Consider

[ \sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^{n} (x+1)^{n}. ]

Here (a_n=\left(\frac{3}{4}\right)^{n}) and the center is (c=-1).
Applying the root test:

[ \sqrt[n]{|a_n|}= \sqrt[n]{\left(\frac{3}{4}\right)^{n}} = \frac{3}{4}, ]

so (L=\frac{3}{4}) and

[ R=\frac{1}{L}= \frac{4}{3}. ]

Thus the series converges when (|x+1|<\frac{4}{3}), i.e. for (x\in\left(-\frac{7}{3},\frac{1}{3}\right)). Endpoint checking (substituting (x=-\frac{7}{3}) and (x=\frac{1}{3})) shows divergence at both ends because the series reduces to (\sum 1^{n}) and (\sum (-1)^{n}), respectively.

Example 4: Using the (\limsup) (Cauchy–Hadamard)

Find the radius for

[ \sum_{n=0}^{\infty} a_n (x-5)^{n},\qquad a_n=\begin{cases} 2^{-n}, & n\text{ even},\[2pt] 3^{-n}, & n\text{ odd}. \end{cases} ]

The ordinary limit (\lim_{n\to\infty}\sqrt[n]{|a_n|}) does not exist because the subsequences give different values ((\frac12) and (\frac13)). Compute the (\limsup):

[ \limsup_{n\to\infty}\sqrt[n]{|a_n|}= \max\left{\frac12,\frac13\right}= \frac12 . ]

Hence

[ \frac{1}{R}= \frac12 \quad\Longrightarrow\quad R=2 . ]

The series converges for (|x-5|<2), i.e. (x\in(3,7)). At the endpoints the series becomes (\sum 2^{-n}) (convergent) for (x=3) and (\sum (-1)^{n}2^{-n}) (also convergent) for (x=7); thus the interval of convergence is actually ([3,7]).

Conclusion

Determining the radius of convergence of a power series reduces to evaluating a limit (or (\limsup)) of the (n)-th root of the coefficient magnitude. The ratio test offers a quick route when factorials or simple ratios appear; the root test excels when coefficients are already expressed as (n)-th powers; and the Cauchy–Hadamard formula guarantees a result even when ordinary limits fail. After obtaining (R), one must still examine the boundary points (x=c\pm R) separately, as convergence there is not decided by the radius test alone. Mastery of these tools enables efficient analysis of power series in both theoretical and applied contexts.

Example 5: Ratio Test with Factorial‑Power Coefficients

Consider

[ \sum_{n=1}^{\infty}\frac{(2n)!}{n!,4^{,n}},x^{n}. ]

Here

[a_n=\frac{(2n)!}{n!,4^{,n}} . ]

Applying the ratio test:

[ \left|\frac{a_{n+1}}{a_n}\right| =\frac{(2n+2)(2n+1)(2n)!}{(n+1)!,4^{,n+1}} \cdot\frac{n!,4^{,n}}{(2n)!} =\frac{(2n+2)(2n+1)}{4(n+1)} =\frac{(2n+2)(2n+1)}{4n+4}. ]

Taking the limit as (n\to\infty),

[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\frac{4n^{2}+6n+2}{4n+4} \longrightarrow \infty . ]

Thus (L=\infty) and

[ R=\frac{1}{L}=0 . ]

The series converges only at the centre (x=0); for any non‑zero (x) the terms do not tend to zero. This illustrates that a radius of zero is admissible and that the ratio test can signal divergence for every point outside the trivial centre.

Example 6: Root Test with Polynomially Decaying Coefficients

Let

[ \sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)^{2}},x^{n}. ]

The coefficients are (a_n=\frac{(-1)^{n}}{(n+1)^{2}}).
The root test gives

[ \sqrt[n]{|a_n|}= \frac{1}{(n+1)^{2/n}} \longrightarrow 1 \qquad (n\to\infty), ]

so (L=1) and consequently (R=1).
The disc of convergence is (|x|<1).
At the boundary points (x=1) and (x=-1) the series becomes (\sum (-1)^{n}/(n+1)^{2}) and (\sum 1/(n+1)^{2}), both of which converge absolutely; therefore the interval of convergence actually closes to ([-1,1]).

Example 7: Power Series in a Transformed Variable

Suppose we wish to study

[ \sum_{n=0}^{\infty}\frac{x^{2n}}{n!}. ]

Writing (y=x^{2}) converts the series into

[ \sum_{n=0}^{\infty}\frac{y^{n}}{n!}, ]

which is the exponential series with radius (R_y=\infty). Hence the original series converges for every real (x); the radius in the (x)-

variable is effectively infinite because (y=x^{2}) is always non-negative and the exponential series has no singularities.

Example 8: Cauchy–Hadamard When Limits Fail

Consider

[ \sum_{n=0}^{\infty}\frac{x^{n}}{n!}. ]

Here (a_n=1/n!). The ratio test yields

[ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{1}{n+1}=0, ]

so (L=0) and (R=\infty). The root test gives the same result:

[ \lim_{n\to\infty}\sqrt[n]{|a_n|} =\lim_{n\to\infty}\frac{1}{(n!)^{1/n}}=0. ]

Both methods agree, but if the coefficients were more irregular, the Cauchy–Hadamard formula

[ \frac{1}{R}=\limsup_{n\to\infty}\sqrt[n]{|a_n|} ]

would still guarantee a well-defined radius.

Conclusion

Determining the radius of convergence of a power series reduces to evaluating a limit (or (\limsup)) of the (n)-th root of the coefficient magnitude. The ratio test offers a quick route when factorials or simple ratios appear; the root test excels when coefficients are already expressed as (n)-th powers; and the Cauchy–Hadamard formula guarantees a result even when ordinary limits fail. After obtaining (R), one must still examine the boundary points (x=c\pm R) separately, as convergence there is not decided by the radius test alone. Mastery of these tools enables efficient analysis of power series in both theoretical and applied contexts.