$\sum_{n=1}^\infty \frac{1}{2^n-1}$

Question

What is the exact value of$$\sum_{n=1}^\infty \frac{1}{2^n-1}$$ if it exists?

Here's a tutorial in MathJax – Sahiba Arora Jun 08 '17 at 17:45 — Sahiba Arora, Jun 08 '17 at 17:45
Why shouldn't the exact value exist? – Jack D'Aurizio Jun 08 '17 at 19:38 — Jack D'Aurizio, Jun 08 '17 at 19:38

score 0 · Answer 1 · answered Jun 08 '17 at 17:31

0

https://www.wolframalpha.com/input/?i=sum+(2%5En-1)%5E-1+from+1+to+infty

It's about 1.6067, no exact value

answered Jun 08 '17 at 17:31

Akababa

3,109

score 0 · Answer 2 · answered Jun 08 '17 at 18:10

0

Hint. $$ \frac{1}{2^n-1}=\frac{1}{2^n}\cdot\frac{1}{1-2^{-n}}=\frac{1}{2^n}\sum_{k=0}^\infty \frac{1}{2^{kn}}=\sum_{k=1}^\infty \frac{1}{2^{kn}}. $$ Hence $$ \sum_{n=1}^\infty\frac{1}{2^n-1}=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{2^{kn}}=\sum_{m=1}^\infty\frac{f(m)}{2^m}, $$ where $f(m)$ is the number of divisors of $m$.

answered Jun 08 '17 at 18:10

Yiorgos S. Smyrlis

83,933

1

that's not really much simpler than the original series... – Albert Jun 08 '17 at 18:31

$\sum_{n=1}^\infty \frac{1}{2^n-1}$

2 Answers2