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What could be an intuitive explanation for $\displaystyle \sum_{k=1}^{\infty}{\frac1{k\,2^k}} = \log 2$ ?

$\displaystyle \sum_{k=1}^{\infty}{\frac{1}{2^k}} = 1$ has a simple intuitive explanation with taking a unit distance and adding the successive halfs consecutively, is there a similar explanation for $\ln 2$?

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Oddly enough someone else just asked this question yesterday, before deleting it. Anyway, try the Taylor expansion of $-\log(1/2)$. – John M Aug 5 '11 at 23:05
In order for their to be an "intuitive" explanation, you'd have to start with an intuitive way to understand $\ln 2$ in the first place. – anon Aug 5 '11 at 23:09
@John: This isn't a coincidence: – t.b. Aug 5 '11 at 23:18
@John I am the culprit, though it was not yesterday but the day before I think. I have posted an answer. – kuch nahi Aug 6 '11 at 1:12
log is the inverse function to taking powers. You sum, more or less, over powers, which is, more or less, like integrating over them, and we know this is related to values of the antiderivative. – Mark Aug 6 '11 at 1:18

For $|t| < 1$, $\sum_{k=0}^\infty t^k = \frac{1}{1-t}$. This is a geometric series, with an "intuitive explanation" similar to yours for $\sum_{k=0}^\infty \frac{1}{2^k}$. Take the derivative of $\sum_{k=1}^\infty \frac{t^k}{k}$ term-by-term and you get $\sum_{k=0}^\infty t^k$.

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I don't know whether you'll find this intuitive, but here is how I think about the Taylor expansion of the logarithm, which I will write as

$$\log \frac{1}{1 - x} = \sum_{k \ge 1} \frac{x^k}{k}.$$

(We get your identity by setting $x = \frac{1}{2}$.) This is equivalent to the identity

$$\frac{1}{1 - x} = \sum_{n \ge 0} x^n = \exp \left( \sum_{k \ge 1} \frac{x^k}{k} \right)$$

which admits the following combinatorial interpretation: the LHS is the exponential generating function of the number of permutations of a set of size $n$, and the RHS, as it turns out, is the generating function for the number of ways you can divide a set of size $n$ up into cycles - but this is exactly the cycle decomposition of a unique permutation!

This argument is explained in more detail in a series of posts on my blog starting here. It is a special case of the exponential formula in combinatorics and can also be proven (as I did in the posts above) using Pólya's enumeration theorem. You may also be interested to learn what this has to do with zeta functions.

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This is a question I asked recently and answered myself, so I deleted it. Though someone had posted the hint in the comment. Here is what I worked out. $$\log 2 = -\log\frac{1}{2} $$

Now you can use the expansion of $\log (1+x)$ $$\log (1+x) = \sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k}\qquad \mbox{for}\;\; |x|<1$$ with $x = -\frac{1}{2}$ to get

$$\log 2 = -\log\frac{1}{2} =\sum_{k=1}^\infty \frac{1}{k2^k}$$

(Although I cannot provide an intuitive explanation, I can give you an application. My motivation for asking the question was that this formula can be used to determine individual binary digits of $\log 2$ and was the starting point for the authors to discover what is now called BBP Formula to determine the $n^{th}$ digit of $\pi$ wihtout having to computing the first $n-1$ digits, in base 2 and 16 )

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For the record, when you answer a question yourself, you should post the answer to your own question. Deleting it doesn't help anybody! :) – Adrian Petrescu Aug 24 '11 at 16:02

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