Does Σ 3n/(n^2 + 4) converge Step by Step

Step by step

The question

Question: decide if the series converges: the sum of 3n/(n² + 4) for n from 1 to infinity.

Σ_{n = 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 a_n → 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.

Answer