Recipe

Find the radius of convergence and the open interval

Use when: a power series sum a_n(x-c)^n is given and the set of x-values where it converges is needed

Goal: the radius R and the open interval (c-R,c+R), with the two endpoints flagged as still needing separate testing

Taught in Radius and interval of convergence · a later section extends this recipe with more cases

1. 1

   Read off the center c from the series' (x-c)^n factor, then compute \displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}(x)}{a_n(x)}\right|, treating x as a fixed but unknown number.

   Example

   \displaystyle\sum_{n=1}^\infty\frac{(x-2)^n}{n\cdot3^n}: center c=2. \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{(x-2)^n}\right| = \frac{n}{n+1}\cdot\frac{|x-2|}{3}

   Why: the Ratio Test's limit is always linear in |x-c|, which is exactly why the convergence set is guaranteed to be an interval centered at c, regardless of what the coefficients a_n are.

2. 2

   Let n\to\infty in the ratio from Step 1; only the |x-c| dependence survives.

   Example

   \dfrac{n}{n+1}\to1, so L(x)=\dfrac{|x-2|}{3}.

   Why: this collapses the ratio down to a single expression in |x-c| alone, ready to compare against the Ratio Test's threshold of 1.

3. 3

   Set the limit from Step 2 less than 1 and solve the resulting absolute-value inequality for |x-c|; the radius R is whatever bound that inequality puts on |x-c|.

   Example

   \dfrac{|x-2|}{3}<1 \implies |x-2|<3, so R=3.

   Why: the Ratio Test converges exactly when L(x)<1, so solving that inequality is the definition of finding R, not just a computational shortcut.

4. 4

   Convert |x-c|<R into the open interval (c-R,c+R). The two endpoints, where L(x)=1 exactly, are not yet resolved by the Ratio Test and need separate testing.

   Example

   \boxed{R=3}, \quad \text{open interval } (-1,5) \text{ so far}

   The endpoints x=-1 and x=5 are where L(x)=1 and still need to be checked one at a time.

   • Not yet: the two endpoints themselves need to be tested one at a time, as ordinary numerical series — covered in Testing the endpoints

   Why: the Ratio Test is silent, by its own hypothesis, exactly at the two points where L(x)=1, so the open interval alone is not the complete answer.