Worked examples
The question
Question: decide if the series converges: the sum of n/(2n + 1) for n from 1 to infinity.
Σn = 1∞n2n + 1 =
1 The term
A series adds up the terms aₙ for ever.
First look at one term, n/(2n + 1).
an = n2n + 1
The divergence test: if the terms do not go to 0, adding them for ever never settles.
The series diverges.
divergence test:
if an does not → 0,
the series diverges
2 The limit
Divide top and bottom by n, the highest power at the bottom.
an = 12 + 1n
Every part with n at the bottom goes to 0.
as n → ∞ = 12 + 0
So the terms go to 1/2.
= 12
3 Decide
The terms do not go to 0.
Adding numbers close to 1/2 for ever never settles: the series diverges.
= diverges
Careful!
1/2 is where each term ends up, not the sum.
The sum grows by about 1/2 every time.
Careful: not the sum is 1 over 2 ✗
4 Check
Let's check with a calculator: the sums of the first 10, 100, 1000 terms are 4.41, 48.86, 498.28.
They keep growing.
Right!
S₁₀ ≈ 4.41, S₁₀₀ ≈ 48.86, S₁₀₀₀ ≈ 498.28
Let's look back.
First: the term.
Then: the limit.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of (n² + 1)/(3n²) for n from 1 to infinity.
Σn = 1∞n2 + 13n2 =
1 The term
A series adds up the terms aₙ for ever.
First look at one term, (n² + 1)/(3n²).
an = n2 + 13n2
The divergence test: if the terms do not go to 0, adding them for ever never settles.
The series diverges.
divergence test:
if an does not → 0,
the series diverges
2 The limit
Divide top and bottom by n to the 2, the highest power at the bottom.
an = 1 + 1n23
Every part with n at the bottom goes to 0.
as n → ∞ = 1 + 03
So the terms go to 1/3.
= 13
3 Decide
The terms do not go to 0.
Adding numbers close to 1/3 for ever never settles: the series diverges.
= diverges
Careful!
1/3 is where each term ends up, not the sum.
The sum grows by about 1/3 every time.
Careful: not the sum is 1 over 3 ✗
4 Check
Let's check with a calculator: the sums of the first 10, 100, 1000 terms are 3.85, 33.88, 333.88.
They keep growing.
Right!
S₁₀ ≈ 3.85, S₁₀₀ ≈ 33.88, S₁₀₀₀ ≈ 333.88
Let's look back.
First: the term.
Then: the limit.
Then: decide.
Last: check.
▶ Watch it solved
The question
Question: decide if the series converges: the sum of ((−1)ⁿ⁺¹2n + 1)/(n + 3) for n from 1 to infinity.
Σn = 1∞(−1)n+12n + 1n + 3 =
1 The term
A series adds up the terms aₙ for ever.
First look at one term, ((−1)ⁿ⁺¹2n + 1)/(n + 3).
an = (−1)n+12n + 1n + 3
The divergence test: if the terms do not go to 0, adding them for ever never settles.
The series diverges.
divergence test:
if an does not → 0,
the series diverges
2 The limit
The size, without the sign.
Divide top and bottom by n, the highest power at the bottom.
|an| = 2 + 1n1 + 3n
Every part with n at the bottom goes to 0.
as n → ∞ = 2 + 01 + 0
So the size goes to 2.
= 2
With the sign, the terms jump between about 2 and minus that.
They do not go to 0.
(−1)n+1: the sign flips
3 Decide
The terms do not go to 0.
Adding numbers close to 2 for ever never settles: the series diverges.
= diverges
Careful!
The limit of the terms tells us only about the terms, not about the sum.
Careful: the limit of the terms is not the sum
4 Check
Let's check with a calculator: the partial sums keep jumping, −0.68, 1.28, −0.7, 1.3.
They never settle.
Right!
S₁₀₀ ≈ −0.68, S₁₀₁ ≈ 1.28, S₁₀₀₀ ≈ −0.7, S₁₀₀₁ ≈ 1.3
Let's look back.
First: the term.
Then: the limit.
Then: decide.
Last: check.
▶ Watch it solved
