Find the sum of the series $\displaystyle\sum_{k=0}^{\infty}\frac{1}{\pi(2k+1)((\pi(2k+1))^2-4L^2)}, L\in (0,1)$ ?
Is it possible to find the sum?
I know $\displaystyle\sum_{k=0}^{\infty}\frac{1}{((\pi(2k+1))^2-4L^2)}= tan (L)/8L$.
I am trying to split $\dfrac{1}{\pi(2k+1)((\pi(2k+1))^2-4L^2)}$ in to two peaces and tried to apply above identity.
But I am not able to split it !
Is there any way to ans this or atleast Is it possible to make some bound for $\dfrac{1}{\pi(2k+1)}$ inside the sum?
I do not know if this is the answer you could expect.
Using partial fraction decomposition $$a_n=\frac{1}{\pi (2 k+1) \left(\pi ^2 (2 k+1)^2-4 L^2\right)}$$ $$a_n=\frac{1}{8 L^2 (\pi (2 k+1)-2 L)}+\frac{1}{8 L^2 (\pi (2 k+1)+2 L)}-\frac{1}{4 \pi (2 k+1) L^2}=b_n+c_n-d_n$$ Now, using generalized harmonic numbers $$\sum_{k=0}^p b_n=\frac{H_{-\frac{L}{\pi }+p+\frac{1}{2}}-H_{-\frac{L}{\pi }-\frac{1}{2}}}{16 \pi L^2}$$ $$\sum_{k=0}^p c_n=\frac{H_{\frac{L}{\pi }+p+\frac{1}{2}}-H_{\frac{L}{\pi }-\frac{1}{2}}}{16 \pi L^2}$$ $$\sum_{k=0}^p d_n=\frac{H_{p+\frac{1}{2}}+2\log (2)}{8 \pi L^2}$$ leading to $$\sum_{k=0}^p a_n=-\frac{-H_{-\frac{L}{\pi }+p+\frac{1}{2}}-H_{\frac{L}{\pi }+p+\frac{1}{2}}+H_{-\frac{L}{\pi }-\frac{1}{2}}+H_{\frac{L}{\pi }-\frac{1}{2}}+2 H_{p+\frac{1}{2}}+4\log (2)}{16 \pi L^2}$$
Using the asymptotics, we get $$\sum_{k=0}^p a_n=-\frac{H_{-\frac{L}{\pi }-\frac{1}{2}}+H_{\frac{L}{\pi }-\frac{1}{2}}+4\log (2)}{16 \pi L^2}-\frac{1}{16 \pi ^3 p^2}+O\left(\frac{1}{p^3}\right)$$ and then the limit you are looking for.