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

Step by step

The question

Question: decide if the series converges: the sum of (2n + 3)/(n³ + 4n) for n from 1 to infinity.

Σ_{n = 1}^∞2n + 3n³ + 4n =

1 For big n

For big n, only the highest powers matter, top and bottom.

a_n = 2n + 3n³ + 4n ≈ 2nn³

Cancel: about 2/n².

= 2n²

2 The known series

We know the sum of 1/n to the 2: a p-series with p = 2.

p is bigger than 1, so it converges.

A number in front, 2, does not change that.

Σ_{n = 1}^∞1n²: p = 2 > 1, converges

3 Compare

The terms are not simply smaller here.

So use the limit comparison test: divide aₙ by bₙ and let n grow.

lim a_nb_n =

Dividing by 2/n² is the same as times its flip.

= n³ + 3n²2n³ + 4n

Top and bottom now have the same highest power with the same number: the limit is 1.

as n → ∞ = 1

The limit is a positive number.

Then the two series do the same thing: both converge, or both diverge.

limit > 0 (a number): both do the same

4 Decide

So our series converges.

= converges

Careful!

Smaller than a series that diverges tells us nothing.

Compare with one that converges.

Careful: not a_n < 1n, so it diverges ✗

5 Check

Let's check with a calculator: 100, 1000 and 10000 terms add up to about 2.2696, 2.2877 and 2.2895.

It settles.

Right!

S₁₀₀ ≈ 2.2696, S₁₀₀₀ ≈ 2.2877, S₁₀₀₀₀ ≈ 2.2895

Let's look back.

First: for big n.

Then: the known series.

Then: compare.

Then: decide.

Last: check.

Answer