Worked examples
The question
Question: decide if the series converges: the sum of 1/(n² + 3) for n from 1 to infinity.
Σn = 1∞1n2 + 3 =
1 For big n
For big n, only the highest powers matter, top and bottom.
an = 1n2 + 3 ≈ 1n2
2 The known series
We know the sum of 1/n to the 2: a p-series with p = 2.
p is bigger than 1, so it converges.
Σn = 1∞1n2: p = 2 > 1, converges
3 Compare
The bottom has extra plus terms: it is bigger than n².
n2 + 3 > n2
A bigger bottom makes a smaller fraction: every term is smaller than 1/n².
an < 1n2
The direct comparison test: our terms are smaller than terms that add up to a number.
So ours add up to a number too.
smaller than a converging series: converges
4 Decide
So our series converges.
= converges
Careful!
Smaller than a series that diverges tells us nothing.
Compare with one that converges.
Careful: not an < 1n, so it diverges ✗
5 Check
Let's check with a calculator: 100, 1000 and 10000 terms add up to about 0.7303, 0.7393 and 0.7402.
It settles.
Right!
S100 ≈ 0.7303, S1000 ≈ 0.7393, S10000 ≈ 0.7402
Let's look back.
First: for big n.
Then: the known series.
Then: compare.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of (n + 1)/(n³ + 2) for n from 1 to infinity.
Σn = 1∞n + 1n3 + 2 =
1 For big n
For big n, only the highest powers matter, top and bottom.
an = n + 1n3 + 2 ≈ nn3
Cancel: about 1/n².
= 1n2
2 The known series
We know the sum of 1/n to the 2: a p-series with p = 2.
p is bigger than 1, so it converges.
Σn = 1∞1n2: p = 2 > 1, converges
3 Compare
The terms are not simply smaller here.
So use the limit comparison test: divide aₙ by bₙ and let n grow.
lim anbn =
Dividing by 1/n² is the same as times its flip.
= n3 + n2n3 + 2
Top and bottom now have the same highest power with the same number: the limit is 1.
as n → ∞ = 1
The limit is a positive number.
Then the two series do the same thing: both converge, or both diverge.
limit > 0 (a number): both do the same
4 Decide
So our series converges.
= converges
Careful!
Smaller than a series that diverges tells us nothing.
Compare with one that converges.
Careful: not an < 1n, so it diverges ✗
5 Check
Let's check with a calculator: 100, 1000 and 10000 terms add up to about 1.4147, 1.4237 and 1.4246.
It settles.
Right!
S100 ≈ 1.4147, S1000 ≈ 1.4237, S10000 ≈ 1.4246
Let's look back.
First: for big n.
Then: the known series.
Then: compare.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of (n² + 1)/(n³ + 3) for n from 1 to infinity.
Σn = 1∞n2 + 1n3 + 3 =
1 For big n
For big n, only the highest powers matter, top and bottom.
an = n2 + 1n3 + 3 ≈ n2n3
Cancel: about 1 over n.
= 1n
2 The known series
We know the sum of 1/n: a p-series with p = 1.
p is 1, so it diverges.
Σn = 1∞1n: p = 1, diverges
3 Compare
The terms are not simply smaller or bigger here.
So use the limit comparison test: divide aₙ by bₙ and let n grow.
lim anbn =
Dividing by 1/n is the same as times its flip.
= n3 + nn3 + 3
Top and bottom now have the same highest power with the same number: the limit is 1.
as n → ∞ = 1
The limit is a positive number.
Then the two series do the same thing: both converge, or both diverge.
limit > 0 (a number): both do the same
4 Decide
So our series diverges.
= diverges
Careful!
Smaller than a series that diverges tells us nothing.
We need bigger than a diverging one, or the limit test.
Careful: not an < 1n, so it converges ✗
5 Check
Let's check with a calculator: 100, 1000 and 10000 terms add up to about 4.66, 6.96 and 9.26.
It keeps growing.
Right!
S100 ≈ 4.6591, S1000 ≈ 6.9573, S10000 ≈ 9.2594
Let's look back.
First: for big n.
Then: the known series.
Then: compare.
Then: decide.
Last: check.
▶ Watch it solved
