Steply

Does Σ 3^n/n^2 converge

Step by step, with the reason for every step · The ratio and root tests · Calculus 2

Step by step

The question

Question: decide if the series converges: the sum of 3ⁿ over n² for n from 1 to infinity.

Σn = 1∞3nn2 =

1 The next term

The ratio test: divide each term by the one before it, and let n grow.

If the limit L is less than 1, the terms shrink like a geometric series: it converges.

Bigger than 1: it diverges.

Exactly 1: the test cannot tell.

ratio test: L = lim an+1an
L < 1 converges, L > 1 diverges

The next term: put n + 1 wherever there is an n.

an+1 = 3n+1(n + 1)2

2 The ratio

Dividing by aₙ is the same as times its flip.

an+1an = 3n+1(n + 1)2 × n23n

Group the same kinds together.

3 to the n + 1, over 3 to the n, is just 3.

= 3 × (nn + 1)2
bn+1 ÷ bn = b

3 The limit

As n grows: n + 1/n goes to 1.

So L = 3.

L = 3

4 Decide

L = 3 is bigger than 1: the series diverges.

= diverges

Careful!

L = 1 tells us nothing.

Only less than 1 converges, and more than 1 diverges.

Careful: not L = 1, so it converges ✗

5 Check

Let's check with a calculator: 5, 10, 20 terms add up to 23.0325, 1023.9208, huge.

It grows without end.

Right!

S5 ≈ 23.0325, S10 ≈ 1023.9208, S20 ≈ huge

Let's look back.

First: the next term.

Then: the ratio.

Then: the limit.

Then: decide.

Last: check.

Answer
= diverges

Your turn

Try this one yourself, then check:

Σn = 1∞4nn2
Show the answer
diverges

More from The ratio and root tests

The whole lesson: The ratio and root tests →