Does Σ 1/n^(3/2) converge Step by Step

Step by step

The question

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

Σ_{n = 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 1n^p 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.

Answer