Consider the following series function defined in terms of an infinite series of the form $$ f(x,y) = \sum_{m=0}^\infty f_m(x,y) , $$ where $$ f_m(x,y) = (2m+2)(2m+3)(2m+5) \, \Gamma \left( m + 7/2 \right)^{-2} \, {}_2F_1 \left( -m, -m-1/2; 2; x^2 \right) y^{2m} \, . $$ Here, $\Gamma$ is the gamma function and ${}_2F_1$ is Gauss hypergeometric function.
To evaluate the function numerically up to a given precision, the series needs to be truncated after a certain number of terms. While for $x \to \infty$ and $m \to \infty$, I was able to show that $$ f_n(x,y) \sim \frac{8}{\pi^{3/2}} \left( \frac{xy \, \mathrm{e}}{m} \right)^{2m} m^{-7/2} \, , $$ I have no idea how to estimate the numbers of terms required to achieve a given precision. Any input is highly appreciated. Thank you!
If you just look at the series $\sum_{m=1}^\infty C (xye/m)^{2m} m^{-7/2}$ (discarding the $m=0$ term since the asymptotic is singular there), you can use the same kind of estimates that are used to prove the ratio test. Specifically, once $(xye/M)<\epsilon<1$ and $M>0$, you have
$$\sum_{m=M}^\infty C (xye/m)^{2m} m^{-7/2}<\sum_{m=M}^\infty C \epsilon^{2m}=C \frac{\epsilon^{2M}}{1-\epsilon^2}.$$