Use the integral test on Σ 1/n^(2/3) Step by Step

Step by step

The question

Question: decide if the series converges: the sum of 1 over (n^(2/3)) for n from 1 to infinity.

Σ_{n = 1}^∞1n^(2/3) =

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^(2/3)

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^(2/3) dx = lim t → ∞ ∫ from 1 to t of 1x^(2/3) dx

The power rule: add 1 to the power, divide by the new power.

= lim t → ∞ [ 3x^(1/3) ] from 1 to t

3 Let t go

Put in t, then take away the same with 1.

= lim t → ∞ (3t^(1/3) − (3))

As t grows, this part grows without end.

3t^(1/3) → ∞

So the area is infinite.

= ∞

4 Decide

The area is infinite, so the series diverges too.

= diverges

This gives a rule for every power of n: the p-series rule.

Here p is 2/3, not bigger than 1.

p-series: Σ 1n^p converges when p > 1

and diverges when p ≤ 1

Careful!

The terms go to 0, but too slowly: the sum still grows for ever.

Careful: not a_n → 0, so it converges ✗

5 Check

Let's check with a calculator: 100 terms add up to about 11.5, 10000 terms to about 62.19.

Slowly, but it keeps growing.

Right!

S₁₀₀ ≈ 11.5, S₁₀₀₀₀ ≈ 62.19

Let's look back.

First: the function.

Then: the integral.

Then: let t go.

Then: decide.

Last: check.

Answer