Prove that at any arbitrarily large $a > 0$ $lim x_{n} = 0$ if $x_{n} = \frac{a^{n}}{n!}$

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Solution

Let, $x_{n} = \frac{a^{n}}{n!}$ , $a > 0$ .
Now, $\frac{x_{n + 1}}{x_{n}} = \frac{a}{n + 1}$ .
$∴ lim n \to \infty$ $\frac{x_{n + 1}}{x_{n}} =$ $0 < 1$ .
By D'Alembert ratio test the series $\infty \sum n = 0 x_{n}$ is convergent.
We have a series is convergent then its limit of $n^{t h}$ term is $0$ .
$\Rightarrow$ $lim n \to \infty x_{n} = 0$ .

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