Steply

Interval of convergence of Σ n (x - 1)^n

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 n (x − 1)ⁿ.

Σn = 1∞n(x − 1)n =

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 − 1) to the n + 1, over (x − 1) to the n, leaves one (x − 1).

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

|an+1an| = |x − 1| × (n + 1n)1

2 The limit

(n + 1)/n goes to 1.

So L is the size of x minus 1.

L = |x − 1|

3 The radius

It converges when L is less than 1.

|x − 1| < 1

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

radius R = 1

Open up the bars: x is between 0 and 2.

0 < x < 2

4 The ends

At x = 2: (x − 1) is 1, so (x − 1) to the n is just 1.

The terms do not go to 0: it diverges.

x = 2: Σn = 1∞n1 : diverges

At x = 0: the same, but with signs that take turns.

The terms do not go to 0: it diverges.

x = 0: Σn = 1∞(−1)nn1 : diverges

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: (0, 2).

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

interval : (0, 2)

6 Check

Let's check an x inside, 1.5, with a calculator: 40 terms and 80 terms add up to almost the same, 2.

It converges there.

Right!

x = 1.5: S40 ≈ 2, S80 ≈ 2

Let's look back.

First: the ratio.

Then: the limit.

Then: the radius.

Then: the ends.

Then: the interval.

Last: check.

Answer
interval : (0, 2)

Your turn

Try this one yourself, then check:

Σn = 1∞n(x − 2)n
Show the answer
(1, 3)

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