I am wondering if there is any closed form of the following summation? $\sum \limits_{i=0}^{\infty} (q^i \sum \limits_{j=0}^i \dfrac{a^j}{j!})$ where |q|<1. I know that $\sum \limits_{i=0}^{\infty} q^i = \dfrac{1}{1-q}$ and also $ \sum \limits_{j=0}^{\infty} \dfrac{a^j}{j!} = e^a $ where |q|<1.
To solve this problem, I think I first need to find the closed form of $ \sum \limits_{j=0}^i \dfrac{a^j}{j!}$ in terms of $a$ and $i$. Then I am going to have only one summation in terms of $i$ (the outer summation), and I can focus on finding the closed form of that.
I hope you can help me to solve this problem.
Change the order of summation: $$ \sum_{i=0}^{\infty} q^i \sum_{j=0}^i \frac{a^j}{j!} = \sum_{j=0}^\infty \frac{a^j}{j!} \sum_{i=j}^\infty q^i = -\frac{1}{q-1}\sum_{j=0}^\infty\frac{(aq)^j}{j!} = -\frac{e^{aq}}{q-1}, $$ for $|q|<1$.