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It isn't hard to prove that:

$$\int_0^x e^{-t} {t^n} dt = n! \cdot e^{-x}\left( e^x-\sum_{k=0}^{n} \frac{x^k}{k!}\right)$$

Or put in a different way:

$$\int_0^x e^{-t} \frac{t^n}{n!} dt = e^{-x}\left( e^x-\sum_{k=0}^{n} \frac{x^k}{k!}\right)$$

Now, if we take the limit as $n\to \infty$ on both sides we will get:

$$\mathop {\lim }\limits_{n \to \infty } \int_0^x e^{-t} \frac{t^n}{n!} dt = \mathop {\lim }\limits_{n \to \infty }e^{-x}\left( e^x-\sum_{k=0}^{n} \frac{x^k}{k!}\right)$$

But $$\mathop {\lim }\limits_{n \to \infty }\sum_{k=0}^{n} \frac{x^k}{k!}= e^x$$

Thus the RHS side is necessarily zero.

This implies that

$$\mathop {\lim }\limits_{n \to \infty } \int\limits_0^x {{e^{ - t}}\frac{{{t^n}}}{{n!}}dt} = 0$$

The idea of the nullity of the limit is that as $n$ increases the "bump" of the function $$y=\displaystyle e^{-x} \frac{x^n}{n!}$$ tends to go further away of the origin, thus for any finite value of $x$ we can choose $N \geq n$ such that the "bump" of the function is sufficiently far from the interval $(0,x)$ that is it made insignificant.

However, I'd like to know if there is a proof for this.

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1 Answer 1

up vote 2 down vote accepted

Hopefully, I miss nothing. I suppose for simplicity that $x\geq 1$. Then $t^n\leq x^n$ for all $t$ in the domain of integration, so $$ \int\limits_0^x e^{-t}\frac{t^n}{n!}dt\leq \int\limits_0^x e^{-t}\frac{x^n}{n!}dt= \frac{x^n}{n!}\int\limits_0^xe^{-t}dt=\frac{x^n}{n!}(1-e^{-x}) $$ and the RHS clearly converges to zero with $n\to\infty$.

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@Peter: sure, done –  Ilya Feb 6 '12 at 17:20
    
@Peter: you're welcome –  Ilya Feb 6 '12 at 17:29

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