<s>[INST] prove $\lim_{n\to\infty}{\frac{a^n}{n!}}=0$ I have the proof but i don't understand one part. The proof (for $a>1$) goes as follows:
there exists $k \in N$ such that$a<k, \frac{a}{k}<1$ and since $\frac{a}{k+i}<\frac{a}{k}, i \in N$
$$\frac{a^n}{n!}=\frac{a}{1} \cdot \frac{a}{2} \cdot...\cdot\frac{a}{k}\cdot\frac{a}{k+1}\cdot...\cdot\frac{a}{n}<\frac{a^k}{k!}(\frac{a}{k})^{n-k}=\frac{a^k}{k!}\cdot \frac{a^na^{-k}}{k^nk^{-k}}$$
$$0<\frac{a^n}{n!}<\frac{k^k}{k!}(\frac{a}{k})^n$$
then $$0 \leq \lim_{n\to\infty}{\frac{a^n}{n!}}\le\frac{k^k}{k!}\lim_{n\to\infty}{(\frac{a}{k})^n}$$
Now, the part I don't understand is in the expansion of $\frac{a^n}{n!}$. How does one come with $<\frac{a^k}{k!}(\frac{a}{k})^{n-k}$ part, to be more precise, how does one come up with $(\frac{a}{k})^{n-k}$ part of the inequality?
 [/INST] We consider $\frac{a}{1}.\frac{a}{2}\ldots \frac{a}{k}\frac{a}{k+1}\ldots \frac{a}{n-2}\frac{a}{n-1}\frac{a}{n}$
Now since $\frac{a}{n}<\frac{a}{n-1}$ and $\frac{a}{n-1}<\frac{a}{n-2}$ and so on till we reach $\frac{a}{n-(n-k)}<\frac{a}{n-(n-k-1)}$ i.e. $\frac{a}{k}<\frac{a}{k-1}$. When we repeatedly use this inequality, we finally will get $\left(\frac{a}{k}\right)^{n-k}$ because there are $n-k$ such terms.
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