From the relation:
$$\csc z=\frac{1}{z}+\sum_{k=1}^\infty (-1)^k \frac{2z}{z^2-k^2 \pi^2}$$
can we obtain the sum of following series?
$$\sum_{k=1}^\infty (-1)^k \frac{2z}{k^2 \pi^2-z^2}\cos kt$$
I have tried using the absolute value, but I got:
$$\sum_{k=1}^\infty \left|\frac{2z}{k^2 \pi^2-z^2}\right|$$
and I can not go on. Any suggestions please?
Consider the series to sum :
$$\tag{1}S(z,t):=\sum_{k=1}^\infty (-1)^k \frac{2z}{k^2 \pi^2-z^2}\cos kt$$
Your first expression is clearly the degenerate case $S(z,0)$.
I'll rather search directly a general expression for $(1)$ using Fourier series for $\cos(xt)$ : $\;\displaystyle \cos(x\,t)=\frac{a_0(x)}2+\sum_{k=1}^{\infty} a_k(x)\cdot \cos(kt)\;$ as exposed here with the result (for $-\pi \le t \le \pi$) :
$$\tag{2}\cos(x\,t)=\frac{2x\sin(\pi x)}{\pi}\left[\frac1{2x^2}+\frac{\cos(1t)}{1^2-x^2}-\frac{\cos(2t)}{2^2-x^2}+\frac{\cos(3t)}{3^2-x^2}-\cdots\right]$$
Set $x:=\dfrac z{\pi}\;$ to get :
$$\tag{3}\frac{\cos\left(z\,t/\pi\right)}{\sin(z)}=2z\left[\frac1{2z^2}+\frac{\cos(1t)}{\pi^2-z^2}-\frac{\cos(2t)}{(2\pi)^2-z^2}+\frac{\cos(3t)}{(3\pi)^2-z^2}-\cdots\right]$$
and the wished expression : $$\tag{4}\sum_{k=1}^\infty (-1)^k \frac{2z}{k^2 \pi^2-z^2}\cos kt=\frac 1z-\frac{\cos\left(z\,t/\pi\right)}{\sin(z)}$$
I didn't use your initial expression (a simple consequence of this) but hope it helped anyway.
