Define the infinite sums $S_1$ and $S_2$ as $$ S_1 = \sum_{m=0}^\infty \frac{\beta^m}{m!}a^{-\frac{m+1}{k}}\Gamma\left(\frac{m+1}{k}\right)$$ and $$S_2 = \sum_{m=0}^\infty (-1)^m \frac{\beta^m}{m!}a^{-\frac{m+1}{k}}\Gamma\left(\frac{m+1}{k}\right) \,.$$ In these equations $\beta,a>0$, $k\geq 3/2$ and $\Gamma$ is the gamma function. Using the ratio test we see that these infinite sums converge for all $k>1$. Accordingly, both $S_1$ and $S_2$ are convergent.
These sums converge slowly, such that machine precision when calculating factorials, exponential or the gamma function becomes an issue.
I want a practical numerical method for calculating these sums such that I can get arbitrarily close to the solution. My attempt consists of truncating the infinite sum. I do, however, need some help in finding an upper bound on the difference between the exact solution and the truncated sum $\|S_{i,\infty} - S_{i, n}\| $. My question is: What are upper error bound estimates for truncated versions of these infinite sums? Other suggestions/alternatives for solving the sums numerically are also welcome.
Note: The issue of machine precision can be solved by representing the infinite sums as the nested functions
$$ S_1 = \beta a^{-\frac{1}{k}}\Gamma\left(\frac{1}{k}\right)\left(1 + \frac{\beta}{1}a^{-\frac{1}{k}}\frac{\Gamma\left(\frac{1}{k}\right)}{B(\frac{2}{k},\frac{1}{k})}\left(1 + \frac{\beta}{2}a^{-\frac{1}{k}}\frac{\Gamma\left(\frac{1}{k}\right)}{B(\frac{2}{k},\frac{1}{k})}\left(1 + ...\right)\right)\right) $$ and $$ S_2 = \beta a^{-\frac{1}{k}}\Gamma\left(\frac{1}{k}\right)\left(1 + \frac{\beta}{1}a^{-\frac{1}{k}}\frac{\Gamma\left(\frac{1}{k}\right)}{B(\frac{2}{k},\frac{1}{k})}\left(1 + \frac{\beta}{2}a^{-\frac{1}{k}}\frac{\Gamma\left(\frac{1}{k}\right)}{B(\frac{2}{k},\frac{1}{k})}\left(1 + ...\right)\right)\right) \,, $$ where $B$ is the beta function.
This is not an answer.
To make things easier, i should try to let $\beta=b^{\frac{1}{k}}$ and then $c=\frac b a$ to make $$S_k = \sum_{m=0}^\infty \frac{\beta^m}{m!}a^{-\frac{m+1}{k}}\Gamma\left(\frac{m+1}{k}\right)=\frac 1{a^{\frac{1}{k}} }\sum_{m=0}^\infty \frac{c^{\frac{m}{k}} }{m!}\Gamma\left(\frac{m+1}{k}\right)$$
If $k$ is a rational number, this seems to be a linear combination of hypergeometric functions for which there are very good subroutines. If $k$ is an integer, $k$ hypergeometric functions should appear.
For example (on purpose, I did not simplify to underline the interesting patterns) $$\color{blue}{a^{\frac 14}\,S_4=\frac{1}{0!}\Gamma \left(\frac{1}{4}\right) \, _0F_2\left(;\frac{2}{4},\frac{3}{4};\frac{c}{4^4}\right)+\frac{1}{1!}\Gamma \left(\frac{2}{4}\right)\, c^{\frac 14} \,\, _0F_2\left(;\frac{3}{4},\frac{5}{4};\frac{c}{4^4}\right)+}$$ $$\color{blue}{\frac{1}{2!} \, \Gamma \left(\frac{3}{4}\right)c^{\frac 12} \, _0F_2\left(;\frac{5}{4},\frac{6}{4};\frac{c}{4^4}\right)+\frac{1}{3!}\Gamma \left(\frac{4}{4}\right) c^{\frac 34} \, _1F_3\left(1;\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{c}{4^4}\right)}$$
$$\color{green}{a^{\frac 15}\,S_5=\frac{1}{0!}\Gamma \left(\frac{1}{5}\right) \, _0F_3\left(;\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{c}{5^5}\right)+\frac{1}{1!} \Gamma \left(\frac{2}{5}\right) c^{\frac 15} \, _0F_3\left(;\frac{3}{5},\frac{4}{5},\frac{6}{5};\frac{c}{5^5}\right)+}$$ $$\color{green}{\frac{1}{2!} \Gamma \left(\frac{3}{5}\right) c^{\frac 25}\, _0F_3\left(;\frac{4}{5},\frac{6}{5},\frac{7}{5};\frac{c}{5^5}\right)+\frac{1}{3!} \Gamma \left(\frac{4}{5}\right) c^{\frac 35}\, _0F_3\left(;\frac{6}{5},\frac{7}{5},\frac{8}{5};\frac{c}{5^5}\right)+}$$ $$\color{green}{\frac{1}{4!}\Gamma \left(\frac{5}{5}\right) c^{\frac 45} \, _1F_4\left(1;\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{c}{5^5} \right)}$$
I am sure that you notice the pattern.
You should not have accuracy problems (at least for this case