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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.

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    $\begingroup$ Try $\sum_{i\ge 0}\sum_{j=0}^if(i,j)=\sum_{j\ge 0}\sum_{i=j}^\infty f(i,j)$ $\endgroup$ – vadim123 Oct 6 '15 at 20:40
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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 $|q|<1$.

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    $\begingroup$ +1 ... as a suggestion, consider stating the basis for the legitimacy of interchanging the order of summations. $\endgroup$ – Mark Viola Oct 6 '15 at 20:56
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    $\begingroup$ @Dr. MV thank you. You can do that because of Fubini's theorem. $\endgroup$ – user153012 Oct 6 '15 at 21:11
  • $\begingroup$ There's an easier way to show this. $\endgroup$ – Mark Viola Oct 6 '15 at 21:19
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    $\begingroup$ Wow! that is really great! thanks a lot. However could you please be so kind to explain how you simplified $\sum_{j=0}^\infty \frac{a^j}{j!} \sum_{i=j}^\infty q^i$ to $ -\frac{1}{q-1}\sum_{j=0}^\infty\frac{(aq)^j}{j!}$ $\endgroup$ – Somi Oct 7 '15 at 21:36
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    $\begingroup$ Hint: $\sum_{i=j}^\infty q^i = -\frac{q^j}{q-1}$. $\endgroup$ – user153012 Oct 7 '15 at 21:49

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