Is the following identity true? 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?
 A: You can evaluate this by using generating functions and integrating.  The answer is $-\pi^2/16 = -0.61685 \ldots$ which is pretty close to $(1-\sqrt{5})/2=-0.61803\ldots$.  
Here's a sketch: the sum is 
$$ 
\sum_{k=1}^{\infty} \frac{(-1)^k}{k} \int_0^1 (1+x^2+ \ldots +x^{2k-2}) dx = \int_0^1 \sum_{j=0}^{\infty} x^{2j} \sum_{k=j+1}^{\infty} \frac{(-1)^k}{k} dx
$$
which is 
$$
 = \int_0^1 \sum_{j=0}^{\infty} x^{2j} \Big(\int_0^1 
\sum_{k=j}^{\infty} -(-y)^{k} dy \Big) dx = - \int_0^1 \int_0^1 \sum_{j=0}^{\infty} \frac{(x^2 y)^j}{1+y} dy dx,
$$
which is 
$$
 = - \int_0^1\int_0^1 \frac{dx dy}{(1+x^2y)(1+y)}.  
$$
The integral in $y$ can be done easily: 
$$ 
\int_0^1 \frac{1}{1-x^2} \Big( \frac{1}{1+y}- \frac{x^2}{1+x^2y}\Big)dy = \log \Big(\frac{2}{1+x^2}\Big) \frac{1}{1-x^2}. 
$$
We're left with 
$$ 
- \int_0^1 \log \frac{2}{1+x^2} \frac{dx}{1-x^2},
$$ 
which WolframAlpha evaluates as $-\pi^2/16$.  (This doesn't look too bad to do by hand, but I don't see a reason to do one variable integrals that a computer can recognize at once.) 
A: 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
$$\sum_{n\geq1}\frac{(-1)^n}{n}\sum_{k=1}^n\frac1{2k-1}=-\frac{\pi^2}{16}.$$
