sum of :$\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$ How can I find the sum of :$$\sum_{k=1}^\infty\frac{(-1)^k}{2k-1} \cos(2k-1)$$
I don't fully understand the parseval identity so I am asking if we can use it to find the sum, and if so, how I should use it.
Is there a Fourier series we know the convergence to a function that can help?

Recall the Maclaurin series of arctangent, valid for $|z|\leq 1,$ $z\neq\pm i$:
$$
\arctan(z) = \sum_{k=0}^{\infty}\frac{(-1)^kz^{2k+1}}{2k+1}
$$
$$
-\arctan(z) = \sum_{k=1}^{\infty}\frac{(-1)^kz^{2k-1}}{2k-1}
$$Put $z=e^{i}$ and take the real part:
$$
\Re(-\arctan(e^i)) = \Re\left( \sum_{k=1}^{\infty}\frac{(-1)^k(e^{i})^{2k-1}}{2k-1}\right)=\sum_{k=1}^{\infty}\frac{(-1)^k\cos(2k-1)}{2k-1}
$$The LHS evaluates to $-1\cdot \pi/4$, since the argument (angle) is $1$ and $\arctan(1)=\pi/4$.
