A power series is a series with a variable in it: $$\begin{aligned} &\sum_{n=0}^{\infty} c_n(x - a)^{n} \\ &= c_0 + c_1(x - a) + c_2(x - a)^{2} + \cdots, \end{aligned}$$ an "infinitely long polynomial" centered at $a$. For each $x$ it is an ordinary series of numbers, so the question becomes: for which $x$ does it converge? The ratio test answers most of that.
Why it works
Apply the ratio test with $x$ held fixed. For many series, $\left|\frac{a_{n+1}}{a_n}\right| \to \frac{|x - a|}{R}$ for some number $R$. The series then converges absolutely when $|x - a| < R$ and diverges when $|x - a| > R$. In general (a theorem worth knowing), every power series converges only at $x = a$, or for all $x$, or exactly inside an interval $|x - a| < R$ and nowhere outside it. At the two endpoints $|x - a| = R$ the ratio test gives $L = 1$: it is silent there.
Key idea
The radius of convergence $R$: the series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$ ($R$ can be 0 or $\infty$). The interval of convergence is $(a - R, a + R)$ plus whichever endpoints converge; test each endpoint separately with a series test.
Worked example: Radius and endpoints
Find the interval of convergence of $\sum_{n=1}^{\infty}\frac{x^{n}}{n}$.
- Ratio: $\left|\frac{x^{n+1}}{n + 1}\cdot\frac{n}{x^{n}}\right| = \frac{n}{n + 1}|x| \to |x|$. Converges for $|x| < 1$: $R = 1$.
- $x = 1$: $\sum\frac{1}{n}$, the harmonic series, diverges.
- $x = -1$: $\sum\frac{(-1)^{n}}{n}$ converges by the alternating series test.
- Interval: $[-1, 1)$.
Answer: $[-1, 1)$
Common mistake
Forgetting to test the endpoints. The ratio test only covers $|x - a| < R$ and $|x - a| > R$; at the endpoints its limit is exactly 1. The two endpoints can behave differently, as in $[-1, 1)$ above. Check: substitute each endpoint into the series and apply a test for number series ($p$-series, alternating series test, divergence test).
Worked example: A shifted center
Find the interval of convergence of $\sum_{n=1}^{\infty}\frac{(x - 3)^{n}}{2^{n}n^{2}}$.
- Ratio: $\frac{|x - 3|}{2}\cdot\frac{n^{2}}{(n + 1)^{2}} \to \frac{|x - 3|}{2}$. Converges when $|x - 3| < 2$: center 3, radius 2.
- $x = 5$: $\sum\frac{1}{n^{2}}$ converges. $x = 1$: $\sum\frac{(-1)^{n}}{n^{2}}$ converges absolutely.
- Interval: $[1, 5]$.
Answer: $[1, 5]$
Worked example: The two extremes
Find the radius of convergence of $\sum\frac{x^{n}}{n!}$ and of $\sum n!\,x^{n}$.
- $\sum\frac{x^{n}}{n!}$: the ratio is $\frac{|x|}{n + 1} \to 0$ for every $x$, so $R = \infty$.
- $\sum n!\,x^{n}$: the ratio is $(n + 1)|x| \to \infty$ unless $x = 0$, so $R = 0$.
Answer: $R = \infty$ and $R = 0$