I found this convergent series while solving a calculus problem $$ \sum_{n=0}^{\infty} \big(e^{(4n+1)\pi\sqrt{3}}+2+e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}=A $$How can I evaluate A?
My Attempt
$$\sum_{n=0}^{\infty} \big(e^{(4n+1)\pi\sqrt{3}}+2+e^{-(4n+1)\pi\sqrt{3}}\big)^{-1}=0.00429610616+1.5281084 \times 10^{-12}+5.3886194 \times 10^{-22}+...$$ The summation appears to converge at around 0.004308 using excel but how can I evaluate this sum analytically?
Consider the partial sum $$S_p=\frac{1}{2}\sum_{n=0}^{+\infty}\frac{1}{1+\text{cosh}[(4n+1)\pi\sqrt{3}]}$$ As mentioned in comments, a CAS gives $$S_p=\frac{\psi _{e^{4 \sqrt{3} \pi }}^{(1)}\left(p-\frac{i}{4 \sqrt{3}}+\frac{5}{4}\right)-\psi _{e^{4 \sqrt{3} \pi }}^{(1)}\left(\frac{1}{12} \left(3-i \sqrt{3}\right)\right)}{48 \pi ^2}$$ where appear the q-polygamma function.
If $p\to \infty$ $$\frac{1}{4 \sqrt{3} \pi }-\frac{\psi _{e^{4 \sqrt{3} \pi }}^{(1)}\left(-\frac{i \pi -\sqrt{3} \pi }{4 \sqrt{3} \pi }\right)}{48 \pi ^2}\approx 0.00429610616779$$
As you noticed, the convergence is very fast $$\left( \begin{array}{cc} p & S_p \\ 0 &\color{red}{ 0.00429610616}62624620267200340161868494728474773150101 \\ 1 &\color{red}{ 0.0042961061677905704599958358}661023829062918667327659 \\ 2 &\color{red}{ 0.0042961061677905704605346978063372}090930922497948456 \\ 3 &\color{red}{ 0.0042961061677905704605346978065272297646}032916399424 \\ 4 &\color{red}{ 0.0042961061677905704605346978065272297646702992589}733 \\ 5 &\color{red}{ 0.0042961061677905704605346978065272297646702992589970} \end{array} \right)$$ which is normal since $$a_n=\frac{1}{1+\text{cosh}[(4n+1)\pi\sqrt{3}]} \sim e^{-\sqrt{3} \pi (4 n+1)}$$ $$\frac{a_{n+1}}{a_n} \sim e^{-4 \sqrt{3} \pi }\approx 3.53 \times 10^{-10}$$