"What is the sum of the series ∑k=1∞−12k−1" was answered by our experts

arrostetzwyl

Answered question

2022-03-14

What is the sum of the series

\sum_{k = 1}^{\infty} - \frac{1}{2^{k} - 1}

Alexa Lewis

Beginner2022-03-15Added 5 answers

Ignoring the negative sign, you're looking for

\frac{1}{2^{1} - 1} + \frac{1}{2^{2} - 1} + \frac{1}{2^{3} - 1} + \dots

but we have

\frac{1}{2^{j k} - 1} = \frac{1}{2^{j k}} + \frac{1}{2^{2 j k}} + \frac{1}{2^{3 j k}} + \dots

If we rewrite the first expression using the second one, then your sum is

(\frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots) + (\frac{1}{2^{2}} + \frac{1}{2^{4}} + \frac{1}{2^{6}} + \dots) + (\frac{1}{2^{3}} + \frac{1}{2^{6}} + \frac{1}{2^{9}} + \dots) + \dots

and

\frac{1}{2^{r}}

appears

τ (r)

times, where

τ (r)

is the number of divisors of r. Therefore your sum is

\sum_{r \geq 1} τ (r) 2^{- r}

and this should allow you to compute it to any desired level of numerical accuracy fairly quickly. For example,

\sum_{r = 1}^{40} τ (r) 2^{- r} = \frac{27602812537}{17179869184} = 1.606695152

and the error here is less than

\sum_{r \geq 41} r 2^{- r} = \frac{84}{2^{41}} < 4 \times 10^{- 11}

since

τ (r) < r

for all

r \geq 3

Of course this method works when 2 is replaced by any constant

c > 1

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