Steply

Does Σ (n + 1)/(n^3 + 2) converge

Step by step, with the reason for every step · Comparison tests · Calculus 2

Step by step

The question

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

Σn = 1∞n + 1n3 + 2 =

1 For big n

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

an = n + 1n3 + 2 ≈ nn3

Cancel: about 1/n².

= 1n2

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.

Σn = 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 anbn =

Dividing by 1/n² is the same as times its flip.

= n3 + n2n3 + 2

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 an < 1n, so it diverges ✗

5 Check

Let's check with a calculator: 100, 1000 and 10000 terms add up to about 1.4147, 1.4237 and 1.4246.

It settles.

Right!

S100 ≈ 1.4147, S1000 ≈ 1.4237, S10000 ≈ 1.4246

Let's look back.

First: for big n.

Then: the known series.

Then: compare.

Then: decide.

Last: check.

Answer
= converges

Your turn

Try this one yourself, then check:

Σn = 1∞n + 14n3 + 1
Show the answer
converges

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