What could be an intuitive explanation for $ \sum\limits_{k=1}^{\infty}\frac{1}{k2^k} = \log 2 $? 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 $\ln 2$?
 A: 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 )
A: 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
$\sum_{k=0}^\infty t^k$.
A: 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. 
