# $\sum\limits_{k=1}^{\infty} {1 \over k}{1 \over 2k-1}$ 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?

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Are there similar series for algebraic multiples of logarithms? – quanta Apr 20 '11 at 14:54

$$\ln2=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}=\sum_{k=1}^\infty\left(\frac{1}{2k-1}-\frac{1}{2k}\right)=\frac{1}{2}\sum_{k=1}^\infty\frac{1}{k}\frac{1}{2k-1}\;.$$

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Ah, that was really a quick answer. And simple, thanks! – Gottfried Helms Apr 20 '11 at 14:45
@GottfriedHelms It's MERCATOR SERIES TIME. – user38268 Apr 21 '11 at 1:01
@David: yes, true. After I asked the question I gave it another try myself, assuming a simple solution based on the mercator series. But Joriki was much quicker and even has it in a one-liner! :-) – Gottfried Helms Apr 21 '11 at 4:54

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 $\ln 2$.

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The downside of this approach is that you're cancelling elements with indices that are apart by a factor of $2$, so you have to make some arguments why you're allowed to do that -- in my approach, you're only combining elements with successive indices, which is more obviously OK. – joriki Apr 20 '11 at 14:45
@joriki, you're right, of course. – lhf Apr 20 '11 at 14:47

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 $$\sum_{k=1}^{\infty} \dfrac1k \dfrac1{2k-1} = 2 \log 2$$

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