Radius and interval of convergence | Code Scholars Tutoring

Understand the idea

A power series is centered at a particular input value. It converges absolutely inside its radius of convergence and diverges outside that radius. The two boundary inputs require separate tests; the radius alone does not say whether either endpoint is included.

A ratio calculation often produces an inequality involving $\displaystyle |x-a|$ . Solve that strict inequality to identify the open interior interval and the radius. A power series can also have radius zero or infinite radius, so do not assume there are always two finite endpoints.

Substitute each endpoint into the original series, including its signs and powers. The resulting numerical series may be harmonic, alternating, geometric, or a p-series. One endpoint can be included while the other is excluded, even though both are the same distance from the center.

\sum c_n(x-a)^n:\quad |x-a|<R\text{ gives absolute convergence; endpoints }a\pm R\text{ require separate tests}

A method you can apply

Apply a suitable convergence test with the variable still present.
Solve the strict inequality and identify the center and radius.
Test each finite endpoint separately and write the final interval with the correct brackets.

Worked example

Find the radius and interval of convergence of $\displaystyle \sum_{n=1}^{\infty}(x-2)^n/(n3^n)$ .

The absolute ratio approaches $\displaystyle |x-2|/3$ , so convergence holds for $\displaystyle |x-2|<3$ .
The center is $\displaystyle 2$ and the radius is $\displaystyle 3$ , giving the interior interval $\displaystyle (-1,5)$ .
At $\displaystyle x=-1$ , the series becomes $\displaystyle \sum(-1)^n/n$ , which converges by the alternating-series test.
At $\displaystyle x=5$ , it becomes the divergent harmonic series.

The two endpoints need different convergence arguments
Endpoint	Resulting series	Conclusion
$\displaystyle x=-1$	$\displaystyle \sum_{n=1}^{\infty}(-1)^n/n$	Converges conditionally
$\displaystyle x=5$	$\displaystyle \sum_{n=1}^{\infty}1/n$	Diverges

The radius is

\displaystyle R=3

and the interval is

\displaystyle [-1,5)

.

Try it yourself

Find the interval of convergence of $\displaystyle \sum_{n=1}^{\infty}(x+1)^n/(n^2 2^n)$ .

Show solution and reasoning

The ratio condition gives $\displaystyle |x+1|<2$ , so the radius is $\displaystyle 2$ and the interior is $\displaystyle (-3,1)$ . At both endpoints, the absolute-value series is $\displaystyle \sum1/n^2$ , which converges. The full interval is $\displaystyle [-3,1]$ .