# Limit of a sequence

Let $$a_n:=\sum_{i=0}^n \frac{n^i}{i!}$$. Find $\lim_{n \to +\infty} \frac{a_n}{e^n}$

Note by Lorenzo Sarnataro
5 years ago

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$$\displaystyle \lim_{n \to \infty} \frac{a_{n}}{e^{n!}}$$

= $$\displaystyle \lim_{n \to \infty} \frac{1 + \frac{n}{1!} + \frac{n^2}{2!} + \dots + \frac{n^{n - 1}}{(n - 1)!} + \frac{n^n}{n!}}{1 + \frac{n}{1!} + \frac{n^2}{2!} + \dots + \frac{n^{n - 1}}{(n - 1)!} + \frac{n^n}{n!} + \frac{n^{n + 1}}{(n + 1)!} + \dots }$$.

Since, $$\dots < \frac{n^{n - 2}}{(n - 2)!} < \frac{n^{n - 1}}{(n -1)!} = \frac{n^n}{n!} > \frac{n^{n + 1}}{(n+1)!} > \frac{n^{n + 2}}{(n+ 2)!} > \dots$$ , we note that denominator is a sequence which increases till $$\frac{n^n}{n!}$$ and then starts decreasing.

Suggest something to be done further.

- 5 years ago

Is the denominator $$e^{n!}$$ or $$e^{n}$$??

- 5 years ago

a(n) is also series expansion of e^n ..... hence following limit becomes 1.

- 5 years ago

I am not of much help here but evaluating this with Wolfram | Alpha gives a very strange answer. Click here.

- 5 years ago

The denominator is $$n!$$, since $$\Gamma(n) = (n-1)!$$ (for integer $$n$$). The numerator is the upper incomplete gamma function, but it doesn't help much with evaluation in any case.

Numerically, the answer appears to be $$1/2$$.

- 5 years ago