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Study the convergence of the series:

$$a_n=\frac{a^2}{n!}\text{ where } a>0$$

So I thought that we should have multiple cases:

Case 1: $a\in (0,1)$

Case 2: $a>1$ but I am not really sure because I know how to find the limit of it and I have done it like that. Can someone help?

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    Series or sequence? – 2017-01-15

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We have $$\lim_{n\to\infty}\frac{a^2}{n!}=0$$ for all $a\in\mathbb{R}$.

Proof. Note that $$\left|\frac{a^2}{n!}\right|\leq\frac{a^2}{n}.$$ Let $\epsilon>0$. Take $N\in\mathbb{N}$ large enough so that $N>a^2/\epsilon$. Then, for all $n\geq N$ we have $a^2/n\leq a^2/N<\epsilon$. $\square$