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The Wikipedia entry on $\ln 2$ includes this summation:

$$\sum_{n=0}^{\infty}{\frac{1}{n \cdot 2^n}} = \ln 2$$

I was very surprised when I saw this, and I don't have a clear sense of where it comes from. This doesn't seem related to the Taylor series for $\ln x$. It somewhat reminds me of the sum

$$\sum_{n=0}^{\infty}{\frac{n}{2^n}}$$

which converges to $2$, but I'm not sure whether there's some way to manipulate the first sum in a way that relates it to the second. It also vaguely reminded me of the sum from the harmonic series, except that the $2^n$ term in the denominator breaks that similarity as well.

Is there a simple explanation as to why this sum gives $\ln 2$? Does this generalize to any other natural logs (say, could a modified sum give back $\ln 3$), or is this oddly specific to $\ln 2$?

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    $\begingroup$ Regarding the last question: We have $\sum_{n=1}^\infty x^n/n = -\ln(1-x)$ for $|x| < 1$. So if you want $\ln y$ you can set $x = 1 - 1/y$. (Although if $y$ is large this will converge slowly.) $\endgroup$
    – Jair Taylor
    Commented May 12, 2020 at 19:14
  • $\begingroup$ Whoops! Missed those earlier questions, though I did so some searching first. :-) Thanks all! $\endgroup$
    – templatetypedef
    Commented May 12, 2020 at 20:15

1 Answer 1

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$ \log(2) = -\log(1/2) = -\log(1 - 1/2) $, and when you use the Taylor series for $ -\log(1-x) $ at $ x = 1/2 $ that's what you get.

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