Power Series and Endpoint Tests

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A series whose input varies

A power series centered at $c$ has form $\sum_{n=0}^\infty a_n(x-c)^n$ . There is a radius $R\in[0,\infty]$ such that it converges absolutely for $|x-c|<R$ and diverges for $|x-c|>R$ . At finite endpoints, each resulting numerical series needs its own test.

The center always converges when its coefficients are finite: every positive-degree term vanishes there. The interval may include both endpoints, one, or neither. The radius alone does not encode this information.

Visual guide

VISUAL GUIDEConvergence changes at the radius

1/(1 − x)
1 + x + ⋯ + x⁶

Finite geometric sums approximate 1/(1 − x) well inside |x| < 1. Near x = 1 the same number of terms is much less accurate. At or outside the radius, the infinite-series conclusion requires a separate check.

Worked example: one endpoint included

Consider $\sum_{n=1}^\infty (x-2)^n/n$ . The ratio of term magnitudes tends to $|x-2|$ , giving radius $1$ and open interval $(1,3)$ .

At $x=1$ , the series is $\sum(-1)^n/n$ , which converges conditionally. At $x=3$ , it is the harmonic series and diverges. Therefore the full interval is $[1,3)$ . The same ratio limit equals $1$ at both endpoints, so the ratio test cannot distinguish them.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

A power series centered at 2 has radius 3. Which points definitely converge without endpoint tests?

−1 ≤ x ≤ 5

−3 < x < 3

−1 < x < 5

Hint 1 · Find a starting point

Use |x−center|<radius.

Hint 2 · Take the next step

The two boundary points require separate investigation.

Show the reasoning

Answer: −1 < x < 5

|x−2|<3 gives (−1,5). Radius alone says nothing decisive about the endpoints.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: all real inputs

For $\sum_{n=0}^\infty x^n/n!$ , the ratio is $|x|/(n+1)\to0$ for every fixed real $x$ . Hence $R=\infty$ . The limit treats $x$ as fixed; it does not say a low-degree truncation is equally accurate for all values of $x$ .

A radius of zero is also possible. For $\sum n!x^n$ , every nonzero $x$ produces a ratio growing without bound, so only $x=0$ converges. Power-series notation by itself does not guarantee a useful interval.

Practice

Find the interval for $\sum_{n=0}^\infty(x/4)^n$ .
Find the interval for $\sum_{n=1}^\infty x^n/n^2$ .
What is the center of $\sum a_n(x+3)^n$ ?

Show worked solutions

$(-4,4)$ . Both endpoints fail the zero-term test.
Radius $1$ ; both endpoints converge absolutely by comparison with $\sum1/n^2$ . Interval $[-1,1]$ .
The center is $-3$ , since $x+3=x-(-3)$ .

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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