I am using it in my m.phil research so it is very vital for me to Evaluating $\sum_{n=0}^\infty\frac1{\Gamma(n/2+2)}$

Question

I want to sum the following: $$\sum_{n=0}^\infty\frac1{\Gamma(n/2+2)}$$

Answer 1

If you are aware of the incomplete gamma function, $$\sum_{x=0}^n \frac{1}{\Gamma \left(\frac{x}{2}+2\right)}=-\left(1+\frac{2}{\sqrt{\pi }}+e\, \text{erfc}(1)\right)+\frac{e \,\Gamma \left(\frac{n+4}{2},1\right)}{\Gamma \left(\frac{n+4}{2}\right)}+\frac{e\, \Gamma \left(\frac{n+3}{2},1\right)}{\Gamma \left(\frac{n+3}{2}\right)}$$ When $n\to \infty$, the sum of the last two terms tend to $2e$ and this makes $$\sum_{x=0}^\infty \frac{1}{\Gamma \left(\frac{x}{2}+2\right)}=e (1+\text{erf}(1))-\left(1+\frac{2}{\sqrt{\pi }}\right)\sim 2.88060$$

Now, you need to prove that this is correct.

Such an answer is of absolutely of no use if you do not know or understand the steps.