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Interval of convergence of Σ (x - 4)^n/5^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 (x − 4)ⁿ over 5ⁿ.

Σn = 1∞(x − 4)n5n =

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 − 4) to the n + 1, over (x − 4) to the n, leaves one (x − 4).

The powers of 5 leave 1/5.

|an+1an| = |x − 4| × 15

2 The limit

So L is 1/5 times the size of x minus 4.

L = 15 × |x − 4|

3 The radius

It converges when L is less than 1.

15 × |x − 4| < 1

Multiply both sides by 5.

|x − 4| < 5

So the radius of convergence is 5: x can be up to 5 away from 4.

radius R = 5

Careful!

Not 1/5!

L is 1/5 times the size, so the size must be less than 5.

Careful: not R = 15 ✗

Open up the bars: x is between −1 and 9.

−1 < x < 9

4 The ends

At x = 9: (x − 4) is 5, and its powers cancel the powers of 5.

The terms do not go to 0: it diverges.

x = 9: Σn = 1∞1 : diverges

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

The terms do not go to 0: it diverges.

x = −1: Σn = 1∞(−1)n : diverges

5 The interval

In interval notation: (−1, 9).

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

interval : (−1, 9)

6 Check

Let's check an x inside, 6.5, with a calculator: 40 terms and 80 terms add up to almost the same, 1.

It converges there.

Right!

x = 6.5: S40 ≈ 1, S80 ≈ 1

Let's look back.

First: the ratio.

Then: the limit.

Then: the radius.

Then: the ends.

Then: the interval.

Last: check.

Answer
interval : (−1, 9)

Your turn

Try this one yourself, then check:

Σn = 1∞(x − 4)n3n
Show the answer
(1, 7)

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