Steply

Does Σ n!/3^n 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 n factorial over 3ⁿ for n from 1 to infinity.

Σn = 1∞n!3n =

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 = (n + 1)!3n+1

2 The ratio

Dividing by aₙ is the same as times its flip.

an+1an = (n + 1)!3n+1 × 3nn!

Group the same kinds together.

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

(n + 1) factorial is n + 1 times n factorial, so their fraction is n + 1.

= 13 × (n + 1)
bn+1 ÷ bn = b
(n + 1)!n! = n + 1

Careful!

(n + 1) factorial is n + 1 times n factorial.

So the fraction is n + 1, not 1.

Careful: not (n + 1)!n! = 1 ✗

3 The limit

As n grows: n + 1 grows without end.

So L = infinity.

L = ∞

4 Decide

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

= diverges

5 Check

Let's check with a calculator: 5, 10, 20 terms add up to 1.5679, 90.8957, huge.

It grows without end.

Right!

S5 ≈ 1.5679, S10 ≈ 90.8957, 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∞n!5
Show the answer
diverges

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