Does $\sum\limits_{n=1}^{\infty}\left(1-e^{-a_n} \sum\limits_{i=0}^{n-1}\frac{(a_n)^i}{i!}\right)$ converge if $(a_n) \subset [0,\infty)$ is bounded? 
Let $(a_n) \subset [0,\infty)$ be bounded. Does $$\sum_{n=1}^{\infty}\left(1-e^{-a_n} \sum_{i=0}^{n-1}\frac{(a_n)^i}{i!}\right)$$
  converge?

I tried rewriting and different things but didn't come to a conclusion.
What I tried:
Let $S_N$ be the $N$'th partial sum of the given series. We have
$$ S_N \leq  \sum_{n=1}^{N}\left(e^{a_n} - \sum_{i=0}^{n-1}\frac{(a_n)^i}{i!}\right). $$
So the question is how fast the inner sum converges to $e^{a_n}$. I guess it's fast enough but I couldn't find a way to show it. 
 A: Hint: Let $0\le a_n\le A$. For $n\ge2A-1$,
$$
\begin{align}
\left|\,e^{a_n}-\sum_{k=0}^{n-1}\frac{a_n^k}{k!}\,\right|
&=\sum_{k=n}^\infty\frac{a_n^k}{k!}\\
&\le\frac{a_n^n}{n!}\sum_{k=n}^\infty\left(\frac{a_n}{n+1}\right)^{k-n}\\
&=\frac{a_n^n}{n!}\frac{n+1}{n+1-a_n}\\[6pt]
&\le2\frac{A^n}{n!}
\end{align}
$$
A: For every $n$, let $X_n$ denote some random variable with Poisson distribution of parameter $a_n$, then the sum under consideration is $$S=\sum_{n=1}^\infty(1-P(X_n\leqslant n-1))=\sum_{n=1}^\infty P(X_n\geqslant n)$$
One further assumes that $(a_n)$ is bounded, say, for every $n$,

$$a_n\leqslant a$$ 

Then every $X_n$ is dominated in distribution by a random variable $X$ with Poisson distribution of parameter $a$, in particular, $$P(X_n\geqslant n)\leqslant P(X\geqslant n)$$ This yields 

$$S\leqslant\sum_{n=1}^\infty P(X\geqslant n)=E(X)=a$$

In particular, $S$ is finite.

Now that probability theory showed the way to a proof, one can, if one wants to, concoct an approach that avoids using explicitely probabilistic notions. Namely, one might want to show that every function $$s_n(x)=e^{-x}\sum_{i=0}^{n-1}\frac{x^i}{i!}$$ is nonincreasing on $x\geqslant0$ because $$s'_n(x)=-e^{-x}\frac{x^{n-1}}{(n-1)!}\leqslant0$$
hence $$s_n(a_n)\geqslant s_n(a)$$ and $$S=\sum_{n=1}^\infty(1-s_n(a_n))\leqslant\sum_{n=1}^\infty(1-s_n(a))=\sum_{n=1}^\infty\sum_{i=n}^\infty e^{-a}\frac{a^i}{i!}=e^{-a}\sum_{i=1}^\infty \frac{a^i}{i!}\sum_{n=1}^i1=e^{-a}\sum_{i=1}^\infty \frac{a^i}{i!}i=\ldots$$ which can be completed easily. Basically, what the proof we presented above accomplishes is to explain each step in this sequence of (otherwise, rather opaque) computations.
