TITLE: Is the following identity true?
QUESTION [44 upvotes]: Calculation suggests the following identity:
$$
\lim_{n\to \infty}\sum_{k=1}^{n}\frac{(-1)^k}{k}\sum_{j=1}^k\frac{1}{2j-1}=\frac{1-\sqrt{5}}{2}.
$$
I have verified this identity for $n$ up to $5000$ via Maple and find that the left-hand side approaches $\frac{1-\sqrt{5}}{2}$. However, this double summation has slow rate of convergence and I am unsure it is true. 
So I want to ask if it is true. If so, how to prove it?

REPLY [12 votes]: Start with
$$\sum_{n\geq1}\frac{z^{2n-1}}{2n-1}=\frac12\log\left(\frac{1+z}{1-z}\right).$$
Since $\frac1{1-z^2}\sum_{n\geq1}\frac{z^{2n-1}}{2n-1}=\sum_{n\geq1}z^{2n-1}\sum_{k=1}^n\frac1{2k-1}$, we have
$$\int_0^z\left(\sum_{n\geq1}z^{2n-1}\sum_{k=1}^n\frac1{2k-1}\right)dz=\int_0^z
\frac1{2(1-z^2)}\log\left(\frac{1+z}{1-z}\right)\,dz.$$
Therefore,
$$\sum_{n\geq1}\frac{z^{2n}}{2n}\sum_{k=1}^n\frac1{2k-1}=
\frac18\log^2\left(\frac{1+z}{1-z}\right).$$
Choose $z=i=\sqrt{-1}$ so that
$$\sum_{n\geq1}\frac{(-1)^n}{2n}\sum_{k=1}^n\frac1{2k-1}=
\frac18\log^2\left(\frac{1+i}{1-i}\right)=-\frac{\pi^2}{32}.$$
It follows that
$$\sum_{n\geq1}\frac{(-1)^n}{n}\sum_{k=1}^n\frac1{2k-1}=-\frac{\pi^2}{16}.$$