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.)

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 $$\sum_{n\geq1}\frac{(-1)^n}{n}\sum_{k=1}^n\frac1{2k-1}=-\frac{\pi^2}{16}.$$