Does Σ (3n^3 - 1)/(n^3 + 4n) converge Step by Step

Step by step

The question

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

Σ_{n = 1}^∞3n³ − 1n³ + 4n =

1 The term

A series adds up the terms aₙ for ever.

First look at one term, (3n³ − 1)/(n³ + 4n).

a_n = 3n³ − 1n³ + 4n

The divergence test: if the terms do not go to 0, adding them for ever never settles.

The series diverges.

divergence test:

if a_n does not → 0,

the series diverges

2 The limit

Divide top and bottom by n to the 3, the highest power at the bottom.

a_n = 3 − 1n³1 + 4n²

Every part with n at the bottom goes to 0.

as n → ∞ = 3 − 01 + 0

So the terms go to 3.

= 3

3 Decide

The terms do not go to 0.

Adding numbers close to 3 for ever never settles: the series diverges.

= diverges

Careful!

3 is where each term ends up, not the sum.

The sum grows by about 3 every time.

Careful: not the sum is 3 ✗

4 Check

Let's check with a calculator: the sums of the first 10, 100, 1000 terms are 22.89, 291.87, 2991.76.

They keep growing.

Right!

S₁₀ ≈ 22.89, S₁₀₀ ≈ 291.87, S₁₀₀₀ ≈ 2991.76

Let's look back.

First: the term.

Then: the limit.

Then: decide.

Last: check.

Answer