Worked examples
The question
Question: decide if the series converges: the sum of 2ⁿ over n factorial for n from 1 to infinity.
Σn = 1∞2nn! =
1 The next term
The ratio test: divide each term by the one before it, and let n grow.
If the limit L is less than 1, the terms shrink like a geometric series: it converges.
Bigger than 1: it diverges.
Exactly 1: the test cannot tell.
ratio test: L = lim an+1an
L < 1 converges, L > 1 diverges
The next term: put n + 1 wherever there is an n.
an+1 = 2n+1(n + 1)!
2 The ratio
Dividing by aₙ is the same as times its flip.
an+1an = 2n+1(n + 1)! × n!2n
Group the same kinds together.
2 to the n + 1, over 2 to the n, is just 2.
(n + 1) factorial is n + 1 times n factorial, so their fraction is 1/n + 1.
= 2 × 1n + 1
bn+1 ÷ bn = b
(n + 1)!n! = n + 1
Careful!
(n + 1) factorial is n + 1 times n factorial.
So the fraction is n + 1, not 1.
Careful: not (n + 1)!n! = 1 ✗
3 The limit
As n grows: 1/n + 1 goes to 0.
So L = 0.
L = 0
4 Decide
L = 0 is less than 1: the series converges.
= converges
5 Check
Let's check with a calculator: 10, 20, 50 terms add up to 6.389, 6.3891, 6.3891.
It settles.
Right!
S10 ≈ 6.389, S20 ≈ 6.3891, S50 ≈ 6.3891
Let's look back.
First: the next term.
Then: the ratio.
Then: the limit.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of n² over 2ⁿ for n from 1 to infinity.
Σn = 1∞n22n =
1 The next term
The ratio test: divide each term by the one before it, and let n grow.
If the limit L is less than 1, the terms shrink like a geometric series: it converges.
Bigger than 1: it diverges.
Exactly 1: the test cannot tell.
ratio test: L = lim an+1an
L < 1 converges, L > 1 diverges
The next term: put n + 1 wherever there is an n.
an+1 = (n + 1)22n+1
2 The ratio
Dividing by aₙ is the same as times its flip.
an+1an = (n + 1)22n+1 × 2nn2
Group the same kinds together.
2 to the n, over 2 to the n + 1, is 1/2.
= 12 × (n + 1n)2
bn+1 ÷ bn = b
3 The limit
As n grows: n + 1/n goes to 1.
So L = 1/2.
L = 12
4 Decide
L = 1/2 is less than 1: the series converges.
= converges
Careful!
L = 1 tells us nothing.
Only less than 1 converges, and more than 1 diverges.
Careful: not L = 1, so it converges ✗
5 Check
Let's check with a calculator: 10, 20, 50 terms add up to 5.8574, 5.9995, 6.
It settles.
Right!
S10 ≈ 5.8574, S20 ≈ 5.9995, S50 ≈ 6
Let's look back.
First: the next term.
Then: the ratio.
Then: the limit.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of n³ over 3ⁿ for n from 1 to infinity.
Σn = 1∞n33n =
1 The next term
The ratio test: divide each term by the one before it, and let n grow.
If the limit L is less than 1, the terms shrink like a geometric series: it converges.
Bigger than 1: it diverges.
Exactly 1: the test cannot tell.
ratio test: L = lim an+1an
L < 1 converges, L > 1 diverges
The next term: put n + 1 wherever there is an n.
an+1 = (n + 1)33n+1
2 The ratio
Dividing by aₙ is the same as times its flip.
an+1an = (n + 1)33n+1 × 3nn3
Group the same kinds together.
3 to the n, over 3 to the n + 1, is 1/3.
= 13 × (n + 1n)3
bn+1 ÷ bn = b
3 The limit
As n grows: n + 1/n goes to 1.
So L = 1/3.
L = 13
4 Decide
L = 1/3 is less than 1: the series converges.
= converges
Careful!
L = 1 tells us nothing.
Only less than 1 converges, and more than 1 diverges.
Careful: not L = 1, so it converges ✗
5 Check
Let's check with a calculator: 10, 20, 50 terms add up to 4.1119, 4.125, 4.125.
It settles.
Right!
S10 ≈ 4.1119, S20 ≈ 4.125, S50 ≈ 4.125
Let's look back.
First: the next term.
Then: the ratio.
Then: the limit.
Then: decide.
Last: check.
▶ Watch it solved
