I wonder if the limit below $$\lim\_{x\rightarrow +\infty} e^{-x}\sum\_{j=0}^{\infty}\frac{x^{j+a}}{\Gamma(j+a+1)}$$
equals 1, for real constant $a>0$, and how shall we get this result?



A limit calculation
If we take $$\chi(x)=\sum\_{j=0}^{\infty} \frac{x^{j+a}}{\Gamma(j+a+1)}$$ then we get, $$\chi'(x)-\chi(x)=\frac{x^{a-1}}{\Gamma(a)}$$


Hence, the solution gives $$\psi(x)=\chi(x)e^{-x}=\frac{1}{\Gamma(a)}{\int e^{-x} x^{a-1} dx}+c$$


This means $$\lim \limits\_{x \to \infty} \psi(x) -\psi(0)=1$$


As, $$\color{grey}{{\int\_{0}^{\infty} e^{-x}x^{a-1} dx=\Gamma(a)}}$$


and $$\color{cadetblue}{\psi(0)=0} $$


So, $$\lim \limits\_{x \to \infty} \psi(x)=e^{-x}\sum\_{j=0}^{\infty} \frac{x^{j+a}}{\Gamma(j+a+1)} =1$$


