Worked examples
The question
Question: find for which x the series converges: the sum of xⁿ over n.
Σn = 1∞xnn =
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)1
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.
That is the sum of 1/n: it diverges.
x = 1: Σn = 1∞1n : diverges
At x = −1: the same, but with signs that take turns.
The signs take turns and the sizes 1/n go to 0: it converges, by the alternating series test.
x = −1: Σn = 1∞(−1)n1n : 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, 0.6931.
It converges there.
Right!
x = 0.5: S40 ≈ 0.6931, S80 ≈ 0.6931
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)
▶ Watch it solved
The question
Question: find for which x the series converges: the sum of (x + 1)ⁿ over (n 2ⁿ).
Σn = 1∞(x + 1)nn 2n =
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 2 leave 1/2.
The powers of n leave (n + 1) over n.
|an+1an| = |x + 1| × 12 × (nn + 1)1
2 The limit
(n + 1)/n goes to 1.
So L is 1/2 times the size of x plus 1.
L = 12 × |x + 1|
3 The radius
It converges when L is less than 1.
12 × |x + 1| < 1
Multiply both sides by 2.
|x + 1| < 2
So the radius of convergence is 2: x can be up to 2 away from −1.
radius R = 2
Careful!
Not 1/2!
L is 1/2 times the size, so the size must be less than 2.
Careful: not R = 12 ✗
Open up the bars: x is between −3 and 1.
−3 < x < 1
4 The ends
At x = 1: (x + 1) is 2, and its powers cancel the powers of 2.
That is the sum of 1/n: it diverges.
x = 1: Σn = 1∞1n : diverges
At x = −3: the same, but with signs that take turns.
The signs take turns and the sizes 1/n go to 0: it converges, by the alternating series test.
x = −3: Σn = 1∞(−1)n1n : converges
5 The interval
In interval notation: [−3, 1).
A square bracket means the end is in; a round one, it is not.
interval : [−3, 1)
6 Check
Let's check an x inside, 0, with a calculator: 40 terms and 80 terms add up to almost the same, 0.6931.
It converges there.
Right!
x = 0: S40 ≈ 0.6931, S80 ≈ 0.6931
Let's look back.
First: the ratio.
Then: the limit.
Then: the radius.
Then: the ends.
Then: the interval.
Last: check.
Answer interval : [−3, 1)
▶ Watch it solved
The question
Question: find for which x the series converges: the sum of n factorial xⁿ.
Σn = 1∞n!xn =
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 factorials leave n + 1.
|an+1an| = |x| × (n + 1)
2 The limit
n + 1 grows without end.
Unless x is 0, L is infinity.
L = ∞ (if x ≠ 0)
3 The interval
The series converges only at x = 0: the radius is 0.
= only x = 0
Careful!
x = 0 still works: every term is 0 there.
Careful: not L = ∞, so no x works ✗
4 Check
Let's check: at x = 0 every term has a 0 in it, so the sum is 0: it converges there.
Right!
x = 0: every term is 0, the sum is 0
Let's look back.
First: the ratio.
Then: the limit.
Then: the interval.
Last: check.
▶ Watch it solved
