# How to calculate the sum of the series $a_n=(c_1+c_2n^2)^{-1.5}$?

Consider the series $a_n=(c_1+c_2n^2)^{-1.5}$ where $n\in\mathbb{N}$.

I want to sum the series up to infinity.

$$(c_1+c_2)^{-1.5}+(c_1+4c_2)^{-1.5}+(c_1+9c_2)^{-1.5}+\cdots$$

I really have no idea because the series is neither in the form of $(n-1)d+a_0$ nor is it like $a_0q^{n-1}$.

How can I get the sum value?

• The general solution depends on the values of $c_1$ and $c_2$ but in the case where $c_1=0$ and $c_2=1$ you get $a_n=\frac{1}{n^3}$ and the sum is $\zeta(3)$ which has no known closed form solution. Aug 2, 2016 at 10:58

Be $0<c_1<c_2$.
$$\sum\limits_{n=1}^\infty \frac{1}{(c_1+c_2 n^2)^{1.5}}=\sum\limits_{n=1}^\infty \frac{1}{c_2^{1.5} n^3}\frac{1}{(1+\frac{c_1}{c_2 n^2})^{1.5}}=\sum\limits_{n=1}^\infty \frac{1}{c_2^{1.5} n^3}\sum\limits_{k=0}^\infty\binom{-1.5}{k}(\frac{c_1}{c_2 n^2})^k$$ $$=\frac{1}{c_2^{1.5}}\sum\limits_{k=0}^\infty\binom{-1.5}{k}(\frac{c_1}{c_2})^k \zeta(2k+3)$$
The values of $\zeta(2k+3)$ give you a possibility for a possible estimation.