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Given any real $a$, I'm trying to prove by the $\epsilon$-$N$ definition that $\frac{a^n}{n!}$ converges. I know how to do this using the ratio test for sequences, but have no luck in using the formal definition, mainly because I cannot bound $n!$.

Any help would be appreciated.

Mathematical Rookie
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    Hint: $a$ is fixed, $n$ is not. Choose $N$ so $n \geq N$ implies $a/n$ is small enough, and use $n(n+1) \cdots (n +k - 1) > n^k$ – tzndls Oct 14 '21 at 01:17
  • Note that $n!\ge (n/2)^{n/2}$. – Mark Viola Oct 14 '21 at 02:32
  • You should be able to get some ideas from the many proofs here: https://math.stackexchange.com/questions/77550/prove-that-lim-limits-n-to-infty-fracxnn-0-x-in-bbb-r – Hans Lundmark Oct 14 '21 at 05:33

2 Answers2

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One way is to notice that if $n > 2a$ then $\dfrac{a}{n} < \dfrac12$.

Then, if $n > 2a$ and $m = \lfloor 2a\rfloor$,

$\begin{array}\\ \dfrac{a^n}{n!} &=\dfrac{\prod_{k=1}^n a}{\prod_{k=1}^n k}\\ &=\dfrac{\prod_{k=1}^m a\prod_{k=m+1}^n a}{\prod_{k=1}^m k\prod_{k=m+1}^n k}\\ &=\dfrac{\prod_{k=1}^m a}{\prod_{k=1}^m k}\dfrac{\prod_{k=m+1}^n a}{\prod_{k=m+1}^n k}\\ &=\prod_{k=1}^m \dfrac{a}{k}\prod_{k=m+1}^n \dfrac{a}{k}\\ &<\dfrac{a^m}{m!}\prod_{k=m+1}^n \dfrac12\\ &=\dfrac{a^m}{m!}\dfrac1{2^{n-m}}\\ &=\dfrac{2^ma^m}{m!}\dfrac1{2^{n}}\\ \end{array} $

For fixed $a$ (and thus $m$), if we choose $n$ large enough so that $\dfrac{2^ma^m}{m!}\dfrac1{2^{n}} \lt \epsilon$, then $\dfrac{a^n}{n!} \lt \epsilon$.

marty cohen
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You can reprove the ratio test for this particular problem:

Let $b_n = \frac{a^n}{n!}$. Let $r_n = |\frac{b_{n + 1}}{b_n}| = \frac{|a|}{n}$. Since $r_n = \frac{|a|}{n} \to 0$ as $n \to \infty$, we can pick $N$ large enough so that $$n \geq N \implies r_n < \frac{1}{2}.$$ We have $$|b_{N + m}| = |b_{N}|r_{N}r_{N + 1}\dots r_{N + m} \leq |b_{N}|\frac{1}{2^{m + 1}} \to 0 \text{ as } m \to 0.$$ Thus $\lim_{m \to \infty}b_m = 0$.

Mason
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