a question about summation of series, how to prove $\int_0^\infty e^{-x}S(x)$=$\sum_{i=0}^\infty a_nn!$ If the coefficients of $\sum_{n=0}^\infty a_nx^n$ is non-negative($a_n\ge 0$ for every n),and the sum function is S(x). Also,suppose$\sum_{i=0}^\infty a_nn!$ is convergent,please prove $\int_0^\infty e^{-x}S(x)$ is also convergent, and it equals to$\sum_{i=0}^\infty a_nn!$. 
Can someone help me solve the question? I totally have no idea how to prove it!i have tried many methods,but I am always stuck.
I will appreciate you very much,if you can help me to find a solution to the question.The fact is that we need to know why $\int_0^\infty e^{-x}S(x)$ is also convergent
 A: This follows from the fact that:
$$
\int_0^\infty e^{-x} x^n = n!
$$
which you can derive by repeatedly integrating by parts. Then apply this formula term-by-term in the series.
A: Since $a_{n}$ are assumed to be non-negative, interchanging the order of integration and summation is justified by Tonelli's theorem, where the convergence issue in the Fubini's Theorem becomes no longer a problem at the expense of imposing non-negativeness condition.
If you are not familiar with this theorem, we may adopt a more problem-specific approach that also generalizes the problem:

Proposition. Assume $a_{n} \in \Bbb{C}$ and $A = \sum_{n=0}^{\infty} a_{n}$ converges. If we define $S(x) = \sum_{n=0}^{\infty} \frac{a_{n}}{n!} x^{n}$, then $\int_{0}^{\infty} S(x)e^{-x} \, dx$ converges and
$$ \int_{0}^{\infty} S(x)e^{-x} \, dx = A. $$ 

Proof. Let $s_{n} = a_{0} + \cdots + a_{n}$ with $s_{-1} = 0$. Then
$$ S(x)
= \sum_{n=0}^{\infty} \frac{s_{n} - s_{n-1}}{n!} x^{n}
= \sum_{n=0}^{\infty} \frac{s_{n}}{n!} x^{n} - \sum_{n=0}^{\infty} \frac{s_{n}}{(n+1)!} x^{n+1} = T'(x) - T(x), $$
where $T(x) := \sum_{n=0}^{\infty} \frac{s_{n-1}}{n!} x^{n} $. So we have
$$ \int_{0}^{R} S(x)e^{-x} \, dx
= \int_{0}^{R} \{ T'(x) - T(x) \} e^{-x} \, dx
= \left[ T(x)e^{-x} \right]_{0}^{R}
= T(R)e^{-R}.
$$
Since $s_{n} \to A$, for any $\epsilon > 0$ we can choose $N$ such that $|s_{n} - A| < \epsilon$ for all $n \geq N$. So
\begin{align*}
\left| T(R)e^{-R} - A \right|
&= e^{-R}\left| T(R) - A e^{R} \right| \\
&\leq e^{-R} \sum_{n=0}^{N} \frac{ \left| s_{n-1} - A \right|}{n!} R^{n} 
      + \epsilon \underbrace{e^{-R}\sum_{n=N}^{\infty} \frac{R^{n}}{n!}}_{\leq 1}.
\end{align*}
Taking limsup to both sides as $R \to \infty$, we have
$$ \limsup_{R\to\infty} \left| T(R)e^{-R} - A \right| \leq \epsilon. $$
But the left-hand side does not depend on $\epsilon$. Thus taking $\epsilon \to 0^{+}$, we have $T(R)e^{-R} \to A$ and the claim follows. ////
