Does Σ 2/(n + 2)^2 converge Step by Step

Step by step

The question

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

Σ_{n = 1}^∞2(n + 2)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) = 2(x + 2)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 2(x + 2)² dx = lim t → ∞ ∫ from 1 to t of 2(x + 2)² dx

The power rule on (x + 2) to the minus 2.

= lim t → ∞ [ −2x + 2 ] from 1 to t

3 Let t go

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

= lim t → ∞ (−2t + 2 − (−23))

As t grows, this part goes to 0.

−2t + 2 → 0

So the area is 2/3: a number.

= 23

4 Decide

The area is finite, so the series converges too.

= converges

Careful!

The area tells us only that the series converges.

The sum itself is a different number.

Careful: not the sum is 2 over 3 ✗

5 Check

Let's check with a calculator: the first 1000 terms add up to about 0.788.

It stays below the first term plus the area, 0.889.

It settles.

Right!

S₁₀₀₀ ≈ 0.788 < a₁ + area = 0.889

Let's look back.

First: the function.

Then: the integral.

Then: let t go.

Then: decide.

Last: check.

Answer