General term formula of $S_n=\sum\limits_{i=1}^n{\frac{a^i}{i!}}$? If $S_n=\sum\limits_{i=1}^n{\frac{a^i}{i!}}$, where a is a positive number, then what's the general term formula of $S_n$?
 A: As noted above,
$$
S_n  = \frac{{e^a }}{{n!}}\int_a^\infty  {x^n e^{ - x} \,dx}  - 1,
$$
which can be written
$$
S_n  = \frac{{e^a }}{{n!}}\Gamma (n + 1,a) - 1,
$$
where $\Gamma(\cdot,\cdot)$ is the upper incomplete gamma function. 
A probabilistic approach. Let $N=\lbrace N_t : t \geq 0 \rbrace$ be a Poisson process with rate $a > 0$. Then
$$
S_n : = \sum\limits_{i = 1}^n {\frac{{a^i }}{{i!}}}  = \sum\limits_{i = 0}^n {\frac{{a^i }}{{i!}}}  - 1 = e^a \sum\limits_{i = 0}^n {\frac{{e^{ - a} a^i }}{{i!}}}  - 1 = e^a {\rm P}(N_1  \le n) - 1.
$$
Now, $N_1 \leq n$ if and only if the time, $Y_{n+1}$, of the $(n+1)$th arrival in the process $N$ is greater than $1$. Since $Y_{n+1}$ is a sum of $n+1$ independent exponential random variables with density function $ae^{-ax}$, $x > 0$, it has gamma probability density function
$$
f_{Y_{n+1}}(x)=x^n e^{-ax} a^{n+1} / n!,\;\; x > 0.
$$
Thus
$$
{\rm P}(N_1  \le n) = {\rm P}(Y_{n+1} > 1) = \int_1^\infty  {f_{Y_{n + 1} } (x)\,dx}  = \frac{{a^{n + 1} }}{{n!}}\int_1^\infty  {x^n e^{ - ax} \,dx}. 
$$
Therefore,
$$
S_n  = e^a \frac{{a^{n + 1} }}{{n!}}\int_1^\infty  {x^n e^{ - ax} \,dx}  - 1 = e^a \frac{{a^{n + 1} }}{{n!}}\int_a^\infty  {\bigg(\frac{x}{a}\bigg)^n e^{ - x} \frac{1}{a}} \,dx - 1 = \frac{{e^a }}{{n!}}\int_a^\infty  {x^n e^{ - x} \,dx}  - 1.
$$
A: For completeness I'll provide a more classical route to proving the formula which shows how one could discover the formula in the first place, since it hasn't been broached in the former question linked by Jerry and Poisson processes might be a little involved for the context of this query in particular. The recursive nature of the upper incomplete gamma function,
$$\Gamma(n+1,a)=\int_a^{\infty}x^ne^{-x}dx,$$
comes about because of how it transforms under by-parts integration. Instead of playing around with $u$ and $v$, I'll just replace the integrand above with an equivalent expression derived from the generic product rule, $nx^{n-1}e^{-x}-(x^ne^{-x})'$, so that by a couple theorems of calculus we have
$$\Gamma(n+1,a)=\int_a^{\infty}nx^{n-1}e^{-x}dx-[x^ne^{-x}]_a^{\infty}$$
$$=n\Gamma(n,a)+a^ne^{-a}$$
$$=n!\Gamma(1,a)+e^{-a}\left(n!a+[n(n-1)\cdots(2)]a^2+\cdots+n(n-1)a^{n-2}+na^{n-1}+a^n\right)$$
Note that $\Gamma(1,a)=\int_a^{\infty}e^{-x}dx=e^{-a}$. Hence if we take the above equation and multiply both sides by the factor $e^a/n!$, we arrive at the desired formula (with the addition of a 1 in front of your original sum):
$$\frac{e^a}{n!}\Gamma(n+1,a)=1+a+\frac{a^2}{2}+\cdots+\frac{a^{n-1}}{(n-1)!}+\frac{a^n}{n!}.$$
A: We could use first-year calculus to get an answer.  By Taylor's Theorem with the integral form of the remainder, we have
$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\int_0^x \frac{e^t(x-t)^n}{n!}dt.$$
Thus 
$$S_n=e^a-1-\int_0^a \frac{e^t(a-t)^n}{n!}dt.$$
The expression for $S_n$ can be transformed in various ways, but will remain stubbornly non-elementary.
A: Somehow, this looks to have been asked before...
Having said that, let this be an addendum to Jacques's and Shai's answers: since the incomplete gamma function and the exponential integral $E_n(z)$ are (essentially) the same function, you can recast the answer in terms of this function:
$$S_n=\frac{e^a a^{n+1}}{n!} E_{-n}(a) - 1$$
