I want to evaluate $$\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{(-1)^{n+k}}{(2k+1)(2n+1)^{2k+1}}$$
From $$\arctan{x}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{2k+1}$$ I get $$\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{(-1)^{n+k}}{(2k+1)(2n+1)^{2k+1}}=\sum_{n=0}^\infty (-1)^n\arctan\left(\frac{1}{2n+1}\right)$$
I try to use telescopic method but it doesn't work.
Thanks in advance.
Let: $$ \alpha(x)=\sum_{k=0}^\infty\arctan\left(\frac{x}{2k+1}\right)(-1)^k $$ Thus $\alpha(0)=0$. Then: $$ \frac{\partial}{\partial x}\alpha(x)=\sum_{k=0}^\infty\frac{(-1)^n}{(2 n+1) \left(\frac{x^2}{(2 n+1)^2}+1\right)}=\frac{\pi}{4\cosh\left(\frac{\pi x}{2}\right)} $$ Integrating both sides we obtain: $$ \alpha(x)=\arctan\left(\tanh\left(\frac{\pi x}{4}\right)\right)+C $$ Substituting $x=0$ we get $C=0$. Thus: $$ \alpha(1)=\arctan\left(\tanh\left(\frac{\pi}{4}\right)\right) $$