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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 $S_n$?

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According to WolframAlpha, it's $ (e^a / n!) (\int_a^{\infty} x^n e^{-x} dx) - 1$. I don't believe that has any simpler form. – anon Jul 11 '11 at 1:47
@anon: Lest the question remain unanswered, perhaps you would care to turn your comment into an answer? Conveniently, the integral even has a name. – André Nicolas Jul 11 '11 at 3:23
... namely the incomplete Gamma function, $\Gamma(n+1,a)$. – Robert Israel Jul 11 '11 at 5:28
I know index variable is arbitrary, but would anone else prefer using "k" instead of "i" as an index variable? – Arjang Jul 11 '11 at 6:56
up vote 4 down vote accepted

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,


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):


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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, $$ S_n = e^a \frac{{a^{n + 1} }}{{n!}}\int_1^\infty {x^n e^{ - ax} \,dx} - 1 = e^a \frac{{a^{n + 1} }}{{n!}}\int_a^\infty {\bigg(\frac{x}{a}\bigg)^n e^{ - x} \frac{1}{a}} \,dx - 1 = \frac{{e^a }}{{n!}}\int_a^\infty {x^n e^{ - x} \,dx} - 1. $$

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+1 - It's always fun to see a solution involving unconventional techniques (such as employing the use of probability theory, more specifically Poisson processes) for problems such as this one. – Max Jul 11 '11 at 7:21

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.

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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:

$$S_n=\frac{e^a a^{n+1}}{n!} E_{-n}(a) - 1$$

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