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?



Is the following identity true?
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.) 


