Worked examples
The question
Question: decide if the series converges: the sum of 1/n² for n from 1 to infinity.
Σn = 1∞1n2 =
1 The function
Make a function from the term: put x in place of n.
The series adds f(1), f(2), f(3) and so on.
f(x) = 1x2
For x from 1 on, it is positive, and it gets smaller: a bigger x makes a bigger bottom.
x ≥ 1: f(x) > 0, getting smaller
The integral test: the series and the area under f from 1 to infinity do the same thing.
Both are finite, or both are infinite.
Σ and ∫ from 1 to ∞: both finite, or both ∞
2 The integral
Infinity is not a number: stop at t, then let t grow, like improper integrals.
∫ from 1 to ∞ of 1x2 dx = lim t → ∞ ∫ from 1 to t of 1x2 dx
The power rule: add 1 to the power, divide by the new power.
= lim t → ∞ [ −1x ] from 1 to t
3 Let t go
Put in t, then take away the same with 1.
= lim t → ∞ (−1t − (−1))
As t grows, this part goes to 0.
−1t → 0
So the area is 1: a number.
= 1
4 Decide
The area is finite, so the series converges too.
= converges
This gives a rule for every power of n: the p-series rule.
Here p is 2, bigger than 1.
p-series: Σ 1np converges when p > 1
and diverges when p ≤ 1
Careful!
The area tells us only that the series converges.
The sum itself is a different number.
Careful: not the sum is 1 ✗
5 Check
Let's check with a calculator: the first 1000 terms add up to about 1.644.
It stays below the first term plus the area, 2.
It settles.
Right!
S1000 ≈ 1.644 < a1 + area = 2
Let's look back.
First: the function.
Then: the integral.
Then: let t go.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of 1 over (n^(3/2)) for n from 1 to infinity.
Σn = 1∞1n^(3/2) =
1 The function
Make a function from the term: put x in place of n.
The series adds f(1), f(2), f(3) and so on.
f(x) = 1x^(3/2)
For x from 1 on, it is positive, and it gets smaller: a bigger x makes a bigger bottom.
x ≥ 1: f(x) > 0, getting smaller
The integral test: the series and the area under f from 1 to infinity do the same thing.
Both are finite, or both are infinite.
Σ and ∫ from 1 to ∞: both finite, or both ∞
2 The integral
Infinity is not a number: stop at t, then let t grow, like improper integrals.
∫ from 1 to ∞ of 1x^(3/2) dx = lim t → ∞ ∫ from 1 to t of 1x^(3/2) dx
The power rule: add 1 to the power, divide by the new power.
= lim t → ∞ [ −2√x ] from 1 to t
3 Let t go
Put in t, then take away the same with 1.
= lim t → ∞ (−2√t − (−2))
As t grows, this part goes to 0.
−2√t → 0
So the area is 2: a number.
= 2
4 Decide
The area is finite, so the series converges too.
= converges
This gives a rule for every power of n: the p-series rule.
Here p is 3/2, bigger than 1.
p-series: Σ 1np converges when p > 1
and diverges when p ≤ 1
Careful!
The area tells us only that the series converges.
The sum itself is a different number.
Careful: not the sum is 2 ✗
5 Check
Let's check with a calculator: the first 1000 terms add up to about 2.549.
It stays below the first term plus the area, 3.
It settles.
Right!
S1000 ≈ 2.549 < a1 + area = 3
Let's look back.
First: the function.
Then: the integral.
Then: let t go.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of 3n/(n² + 4) for n from 1 to infinity.
Σn = 1∞3nn2 + 4 =
1 The function
Make a function from the term: put x in place of n.
The series adds f(1), f(2), f(3) and so on.
f(x) = 3xx2 + 4
For x from 1 on, it is positive.
And it gets smaller: the bottom grows faster than the top.
x ≥ 1: f(x) > 0, getting smaller
The integral test: the series and the area under f from 1 to infinity do the same thing.
Both are finite, or both are infinite.
Σ and ∫ from 1 to ∞: both finite, or both ∞
2 The integral
Infinity is not a number: stop at t, then let t grow, like improper integrals.
∫ from 1 to ∞ of 3xx2 + 4 dx = lim t → ∞ ∫ from 1 to t of 3xx2 + 4 dx
u-substitution: u = x² + 4.
Its derivative 2x dx is there, up to the number 3/2.
u = x² + 4, du = 2x dx
The integral is a log.
= lim t → ∞ [ 32 ln(x² + 4) ] from 1 to t
3 Let t go
Put in t, then take away the same with 1.
= lim t → ∞ (32 ln(t² + 4) − 32 ln 5)
As t grows, this part grows without end.
32 ln(t² + 4) → ∞
So the area is infinite.
= ∞
4 Decide
The area is infinite, so the series diverges too.
= diverges
Careful!
The terms go to 0, but too slowly: the sum still grows for ever.
Careful: not an → 0, so it converges ✗
5 Check
Let's check with a calculator: 100 terms add up to about 11.69, 10000 terms to about 25.49.
Slowly, but it keeps growing.
Right!
S₁₀₀ ≈ 11.69, S₁₀₀₀₀ ≈ 25.49
Let's look back.
First: the function.
Then: the integral.
Then: let t go.
Then: decide.
Last: check.
▶ Watch it solved
