Steply

Interval of convergence of Σ x^n/n^2

Step by step, with the reason for every step · Power series · Calculus 2

Step by step

The question

Question: find for which x the series converges: the sum of xⁿ over n².

Σn = 1∞xnn2 =

1 The ratio

Use the ratio test.

x can be negative, so we take sizes, with bars.

ratio test: L = lim |an+1an| < 1

Divide the next term by this one.

x to the n + 1, over x to the n, leaves one x.

The powers of n leave (n + 1) over n.

|an+1an| = |x| × (nn + 1)2

2 The limit

(n + 1)/n goes to 1.

So L is the size of x.

L = |x|

3 The radius

It converges when L is less than 1.

|x| < 1

So the radius of convergence is 1: x can be up to 1 away from 0.

radius R = 1

Open up the bars: x is between −1 and 1.

−1 < x < 1

4 The ends

At x = 1: x is 1, so x to the n is just 1.

A p-series with p = 2: it converges.

x = 1: Σn = 1∞1n2 : converges

At x = −1: the same, but with signs that take turns.

A p-series with p = 2: it converges, even with the signs.

x = −1: Σn = 1∞(−1)n1n2 : converges

Careful!

The ratio test says nothing at the ends, where L is exactly 1.

Each end must be tested on its own.

Careful: test each end on its own

5 The interval

In interval notation: [−1, 1].

A square bracket means the end is in; a round one, it is not.

interval : [−1, 1]

6 Check

Let's check an x inside, 0.5, with a calculator: 40 terms and 80 terms add up to almost the same, 0.5822.

It converges there.

Right!

x = 0.5: S40 ≈ 0.5822, S80 ≈ 0.5822

Let's look back.

First: the ratio.

Then: the limit.

Then: the radius.

Then: the ends.

Then: the interval.

Last: check.

Answer
interval : [−1, 1]

Your turn

Try this one yourself, then check:

Σn = 1∞(x + 3)nn2
Show the answer
[−4, −2]

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