Worked examples
The question
Question: decide if the series converges: the sum of (−1)ⁿ⁺¹ over n for n from 1 to infinity.
Σn = 1∞(−1)n+1n =
1 The size
Write the first terms: the signs take turns.
terms = 1, −0.5, 0.333, −0.25, ...
Take the size of each term, without the sign: b n = 1 over n.
bn = 1n
The alternating series test: if the sizes get smaller and smaller and go to 0, the series converges.
alternating series test:
if bn gets smaller and bn → 0,
the series converges
2 Getting smaller?
n + 1 is bigger than n, so the bottom gets bigger: each size is smaller than the one before.
bn + 1 < bn
3 Going to 0?
The bottom grows without end: the sizes go to 0.
as n → ∞ = 1n → 0
4 Decide
Both checks pass: the series converges, by the alternating series test.
= converges
5 Absolutely or conditionally?
Now drop the signs: the sum of 1/n is a p-series with p = 1.
p is not bigger than 1: it diverges.
Σn = 1∞1n : p-series, p = 1
The series converges only because the signs take turns.
That is called converging conditionally.
only with the signs: conditionally
So the series converges conditionally.
= converges conditionally
Careful!
The signs make a difference!
Without them it diverges, but with them taking turns, it converges.
Careful: not Σ 1n diverges, so this diverges ✗
6 Check
Let's check with a calculator: the sum of 100 terms is about 0.6882, of 101 terms about 0.6981.
They jump less and less, and the sum is between them.
It settles.
Right!
S100 ≈ 0.6882, S101 ≈ 0.6981
Let's look back.
First: the size.
Then: getting smaller?.
Then: going to 0?.
Then: decide.
Then: absolutely or conditionally?.
Last: check.
Answer = converges conditionally
▶ Watch it solved
The question
Question: decide if the series converges: the sum of (−1)ⁿ⁺¹n/(n² + 1) for n from 1 to infinity.
Σn = 1∞(−1)n+1nn2 + 1 =
1 The size
Write the first terms: the signs take turns.
terms = 0.5, −0.4, 0.3, −0.235, ...
Take the size of each term, without the sign: b n = n/(n² + 1).
bn = nn2 + 1
The alternating series test: if the sizes get smaller and smaller and go to 0, the series converges.
alternating series test:
if bn gets smaller and bn → 0,
the series converges
2 Getting smaller?
The bottom grows faster than the top.
The sizes 0.5, 0.4, 0.3 get smaller.
bn + 1 < bn
3 Going to 0?
Divide top and bottom by n to the 2.
bn = 1n1 + 1n2
Every part with n at the bottom goes to 0.
So the sizes go to 0.
as n → ∞ = 01 + 0 = 0
4 Decide
Both checks pass: the series converges, by the alternating series test.
= converges
5 Absolutely or conditionally?
Now drop the signs.
For big n, n/(n² + 1) is like 1/n: that diverges, like the sum of 1 over n.
Σn = 1∞nn2 + 1 : ≈ Σ 1n , p = 1
The series converges only because the signs take turns.
That is called converging conditionally.
only with the signs: conditionally
So the series converges conditionally.
= converges conditionally
Careful!
The signs make a difference!
Without them it diverges, but with them taking turns, it converges.
Careful: not Σ nn2 + 1 diverges, so this diverges ✗
6 Check
Let's check with a calculator: the sum of 100 terms is about 0.2646, of 101 terms about 0.2745.
They jump less and less, and the sum is between them.
It settles.
Right!
S100 ≈ 0.2646, S101 ≈ 0.2745
Let's look back.
First: the size.
Then: getting smaller?.
Then: going to 0?.
Then: decide.
Then: absolutely or conditionally?.
Last: check.
Answer = converges conditionally
▶ Watch it solved
The question
Question: decide if the series converges: the sum of (−1)ⁿn/(n³ + 1) for n from 1 to infinity.
Σn = 1∞(−1)nnn3 + 1 =
1 The size
Write the first terms: the signs take turns.
terms = −0.5, 0.222, −0.107, 0.062, ...
Take the size of each term, without the sign: b n = n/(n³ + 1).
bn = nn3 + 1
The alternating series test: if the sizes get smaller and smaller and go to 0, the series converges.
alternating series test:
if bn gets smaller and bn → 0,
the series converges
2 Getting smaller?
The bottom grows faster than the top.
The sizes 0.5, 0.222, 0.107 get smaller.
bn + 1 < bn
3 Going to 0?
Divide top and bottom by n to the 3.
bn = 1n21 + 1n3
Every part with n at the bottom goes to 0.
So the sizes go to 0.
as n → ∞ = 01 + 0 = 0
4 Decide
Both checks pass: the series converges, by the alternating series test.
= converges
5 Absolutely or conditionally?
Now drop the signs.
For big n, n/(n³ + 1) is like 1/n to the 2: that converges.
Σn = 1∞nn3 + 1 : ≈ Σ 1n2 , p = 2
The series converges even without its signs.
That is called converging absolutely.
Σ bₙ converges too: absolutely
So the series converges absolutely.
= converges absolutely
Careful!
Absolutely means it would converge even with all the signs plus.
Careful: absolutely: it converges even without the signs
6 Check
Let's check with a calculator: the sum of 100 terms is about −0.347, of 101 terms about −0.3471.
They jump less and less, and the sum is between them.
It settles.
Right!
S100 ≈ −0.347, S101 ≈ −0.3471
Let's look back.
First: the size.
Then: getting smaller?.
Then: going to 0?.
Then: decide.
Then: absolutely or conditionally?.
Last: check.
Answer = converges absolutely
▶ Watch it solved
