Does Σ 1/(2^n - 1) converge Step by Step

Step by step

The question

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

Σ_{n = 1}^∞12ⁿ − 1 =

1 For big n

For big n, 2 to the n is huge: the − 1 hardly matters.

So the terms are about 1/2ⁿ.

a_n = 12ⁿ − 1 ≈ 12ⁿ

2 The known series

The sum of 1/2ⁿ is a geometric series: each term is 1/2 of the one before.

r is less than 1, so it converges.

Σ_{n = 1}^∞12ⁿ: r = 12 < 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 =

Divide: 2 to the n/2 to the n minus 1.

= 2ⁿ2ⁿ − 1

The − 1 hardly matters for big n: 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 1.6067, 1.6067 and 1.6067.

It settles.

Right!

S₁₀₀ ≈ 1.6067, S₁₀₀₀ ≈ 1.6067, S₁₀₀₀₀ ≈ 1.6067

Let's look back.

First: for big n.

Then: the known series.

Then: compare.

Then: decide.

Last: check.

Answer