$\sum\limits_{k=1}^{\infty} {1 \over k}{1 \over 2k-1}$ how to show that this is $ 2 \ln 2 $? Maybe I'm just blocked currently, I think I had it done myself some weeks ago but cannot find/recover the derivation of this equality:
$$\sum_{k=1}^{\infty} {1 \over k}{1 \over 2k-1}  =  2 \ln 2 $$
I have the result in my sketchpad and just checked at Wolfram Alpha, that it is correct. But I cannot remember how I did find it; the numerical approximation needs much more terms than I would use normally for a heuristic. So maybe I've taken this from Wolfram Alpha from the beginning; but anyway: I think the derivation cannot be too difficult. I'm stuck at the moment - could someone help with the derivation?
 A: Expand ${1 \over k}{1 \over 2k-1}$ in partial fractions to ${2 \over 2k-1}-{1 \over k}$. Do some cancellation and end up with twice the alternating harmonic series, which converges to $\ln 2$.
A: $$\ln2=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\sum_{k=1}^\infty\left(\frac{1}{2k-1}-\frac{1}{2k}\right)=\frac{1}{2}\sum_{k=1}^\infty\frac{1}{k}\frac{1}{2k-1}\;.$$
A: Joriki has given an excellent answer. Here is another way to do it.
Consider $$f(x,y) = \sum_{k=1}^{\infty} x^{k-1} y^{2k-2} \tag{$\clubsuit$}$$ We then have from the formula for geometric series$$f(x,y) = \dfrac1{1-xy^2} \tag{$\spadesuit$}$$
Now integrate $(\clubsuit)$ and $(\spadesuit)$ to conclude what you want. Below are the details.

Integrate $(\clubsuit)$. We then have
\begin{align}
\int_0^1\int_0^1 f(x,y) dx dy & = \int_0^1 \int_0^1 \sum_{k=1}^{\infty} x^{k-1} y^{2k-2} dx dy\\
& = \sum_{k=1}^{\infty}\int_0^1 \int_0^1 x^{k-1} y^{2k-2} dx dy\\
& = \sum_{k=1}^{\infty} \dfrac1k \dfrac1{2k-1} \tag{$\diamondsuit$}
\end{align}
The change of infinite sum and integration is justified since the convergence of the series is uniform in any closed set of the form $[0,1-\epsilon_1] \times [0,1-\epsilon_2]$.
Now integrate $(\spadesuit)$. We have
\begin{align}
\int_0^1\int_0^1 f(x,y) dx dy & =\int_0^1 \int_0^1 \left(\dfrac{dx}{1-xy^2}\right)dy\\
& = -\int_0^1 \dfrac{\log(1-y^2)}{y^2} dy\\
& = \log 4 \tag{$\heartsuit$}
\end{align}
where we make use of the fact that $$\int \dfrac{dx}{1-xy^2} = - \dfrac{\log(1-y^2)}{y^2} + \text{constant}$$ and $$- \dfrac{\log(1-y^2)}{y^2} = \dfrac{\log(1-x^2)}{x} + \log(1+x) - \log(1-x) + \text{constant}$$
Now comparing $(\diamondsuit)$ and $(\heartsuit)$, we get that
$$\sum_{k=1}^{\infty} \dfrac1k \dfrac1{2k-1} = 2 \log 2$$
