I need help finding a closed-form solution to $\sum_{n=1}^{\infty}\frac{1}{\left(n^{2}+p\right)^{\frac{3}{2}}}$
This came up when I was doing some physics, and I could not find any solution. I do not have enough education to solve it myself and analysis with Wolfram alpha yeilded no results.
As said, no closed form can be expected.
But since this seems to be for physics, I think that we could have rather good approximations of the summation using the simplistic and totally empirical model $$S_p=\sum_{n=1}^{\infty}\frac{1}{\left(n^{2}+p\right)^{\frac{3}{2}}}\sim \frac {a+b\,p}{1+c\, p}$$ Fitted over the range $0 \leq p \leq 100$, with $R^2=0.999976$ we should have $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 1.20359 & 0.00360 & \{1.19645,1.21074\} \\ b & 0.00148 & 0.00008 & \{0.00132,0.00164\} \\ c & 1.35552 & 0.00587 & \{1.34387,1.36717\} \\ \end{array}$$ A better approximation $(R^2 > 0.999999)$ would be $$S_p=\sum_{n=1}^{\infty}\frac{1}{\left(n^{2}+p\right)^{\frac{3}{2}}}\sim \frac {a+b\,p}{1+c\, p+d \, p^2}$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 1.25144 & 0.00039 & \{1.25065,1.25222\} \\ b & 0.08231 & 0.00083 & \{0.08067,0.08395\} \\ c & 1.51799 & 0.00141 & \{1.51520,1.52078\} \\ d & 0.08470 & 0.00088 & \{0.08296,0.08644\} \\ \end{array}$$