Why is the sum of 1/(n2^n) from 1 to infinity equal to log 2?

[subsonicman]

I was looking at this topic: http://mathoverflow.net/questions/1...country-in-which-people-only-want-boys-closed

And the top answer uses the fact that the sum from 1 to infinity of 1/(x2^x) is log 2. Why is this true?

Thanks in advance.

 

[Mark44]

I was looking at this topic: http://mathoverflow.net/questions/1...country-in-which-people-only-want-boys-closed

And the top answer uses the fact that the sum from 1 to infinity of 1/(x2^x) is log 2. Why is this true?

For starters, it's not true. What you wrote is different from what you linked to. This is what's in that answer.

$$ \sum_{n = 0}^{\infty}\frac{1}{2^{n+1}}\frac{n}{n+1} = 1 - log(2)$$

 

[subsonicman]

I just rewrote the thing in the link to only include the part I was having difficulty with.

$$ \sum_{n = 0}^{\infty}\frac{1}{2^{n+1}}\frac{n}{n+1} = \sum_{n = 0}^{\infty}\frac{1}{2^{n+1}}\left(1 - \frac{1}{n+1}\right)=1-\sum_{n=0}^{\infty}\frac{1}{(n+1)2^{n+1}}=1-\sum_{n=1}^{\infty}\frac{1}{n2^{n}}$$. The link says that this is equal to $$1-\textstyle\log 2$$ (or less ambiguously, ln 2) which is why I said what I did in the first post. And I'm pretty sure it is true. I wrote a program and I summed from 1 to 100 and I got almost exactly ln 2.

 

[deluks917]

http://www.ams.org/journals/mcom/1997-66-218/S0025-5718-97-00856-9/S0025-5718-97-00856-9.pdf

Seems to derive this.

 

[micromass]

Consider the geometric series:

\sum_{n=0}^{+\infty} x^n = \frac{1}{1-x}

Then integrating yields

\int_0^t \sum_{n=0}^{+\infty} x^n dx = \sum_{n=0}^{+\infty} \int_0^t x^n dx = \sum_{n=0}^{+\infty} \frac{1}{n+1} t^{n+1} = \sum_{n=1}^{+\infty} \frac{t^n}{n}

And thus

\sum_{n=1}^{+\infty} \frac{t^n}{n} = \int_0^t\frac{1}{1-x} dx = -\log|1-t|

Filling in $t=1/2$ gives us

\sum_{n=1}^{+\infty} \frac{1}{n2^n} = - \log(1/2) = \log(2)

I leave it up to you to justify each step in this calculation rigorously.