# How to show $\lim_{n\to \infty} \frac{x^n}{n!}=0$ [duplicate]

I know how to show $$\lim_{n\to \infty} \frac{x^n}{n!}=0$$ from the perspective of series.

Namely use the ratio test to show $$\sum\frac{x^n}{n!}$$ is convergent, then the nth term must tends to 0. But I remember I have seen another way to do it in the forum.

I think this question may duplicate, but the 'similar question' doesn't show. If it does duplicate, I apologize.

If $$x = 0$$, the result clearly holds. Suppose $$x \neq 0$$. Let $$a_n = x^n / n!$$. Observe that $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|}{n+1},$$ which converges to zero as $$n \to \infty$$. Therefore there's some $$N$$ such that $$\left|\frac{a_{n+1}}{a_n}\right| < \frac{1}{2}$$ for all $$n \geq N$$. This implies that \begin{aligned} |a_{N+1}| &< \frac{1}{2} |a_N|,\\ |a_{N+2}| &< \frac{1}{4} |a_N|,\\ \cdots \\ |a_{N+k}| &< \frac{1}{2^k}|a_N| \end{aligned} Since the RHS converges to zero as $$k \to \infty$$, so does the LHS, and this implies the desired conclusion that $$a_n \to 0$$.

Let $$N \in \mathbb{N}$$ fulfil $$M := \frac{\lvert x \rvert}{N} < 1$$. Then, for $$n>>N$$: $$\frac{\lvert x \rvert^n}{n!} = \frac{\lvert x \rvert^N}{N!} \cdot \prod_{j = N+1}^n \frac{\lvert x \rvert}{j} < \frac{\lvert x \rvert^N}{N!} \cdot \prod_{j = N+1}^n M = \frac{\lvert x \rvert^N}{N!} \cdot M^{n-N+1} \overset{n \rightarrow \infty}{\longrightarrow} 0$$ Therefore also $$\frac{x ^n}{n!} \overset{n \rightarrow \infty}{\longrightarrow} 0.$$

Let $$N\in\mathbb N$$ be such that $$N>|x|$$.

Then, for $$n>N$$, we have

\begin{align} \left|\frac{x^n}{n!}\right| &= \frac{|x|^N}{N!}\cdot\frac{|x|^{n-N}}{(N+1)\cdot(N+2)\cdots (n-1)\cdot n}\\ &\leq \frac{|x|^N}{N!}\cdot \frac{|x|^{n-N}}{(N+1)^{n-N}}\\ & = \frac{|x|^N}{N!}\cdot \left(\frac{|x|}{N+1}\right)^{-N}\cdot\left(\frac{|x|}{N+1}\right)^{n}\\ &=C\cdot q^n\end{align}

where $$C$$ is a constant equal to $$\frac{|x|^N}{N!}\cdot \left(\frac{|x|}{N+1}\right)^{-N}$$ and $$q=\frac{|x|}{N+1}$$ is a positive number smaller than $$1$$.

Assume $$x<4$$. We have, for $$n\ge 4$$

$$\frac{x^n}{n!}=\frac{x^n}{4!\cdot5\cdot6\cdots n}<\frac{x^4\cdot x^{n-4}}{4!\cdot4\cdot4\cdots 4}=\frac{x^4}{4!}\left(\frac x4\right)^{n-4}.$$

The final expression is a geometric progression of common ratio $$<1$$.

For larger $$x$$, you can replace $$4$$ by some larger integer.

$$\frac{|a_{n+1}|}{|a_n|}=\frac{|x|}{n+1}$$ implies that the sequenses $$\{|a_n|\}$$ is decreasing when $$n$$ large enough. Monotone and boundedness theorem implies that the limit $$\lim_{n\to\infty}|x_n|=:a$$ exists. $$|a_{n+1}|=|x_n|\cdot\frac{|x|}{n+1}$$ let $$n\to\infty$$ on both sides, we can get $$a=a\cdot 0.$$ So $$\lim_{n\to\infty}x_n=0\iff\lim_{n\to\infty}|x_n|=0.$$

The factorial can be squeezed this way (link to proof)

$$\left(\frac n3\right)^n

Therefore $$\left|\dfrac{x^n}{n!}\right|<\left|\dfrac{3x}n\right|^n

$$3x$$ being a fixed quantity then for large $$n$$ we have $$0\le\dfrac{3|x|}n

But the inequalities given above are not so straightforward to prove, fortunately for the job at hand, some more basic ones are sufficent (link to proof)

$$\left(\frac n2\right)^\frac n2

It results that $$\left|\dfrac{x^n}{n!}\right|<\left|\dfrac{2x^2}n\right|^\frac n2

For the same reason than previously, $$2x^2$$ being a fixed quantity.

I think it's good to be aware of and to remember these inequalities, they will serve you much as soon as some factorial are involved, and I've seen many questions on this site that can be solved right away while using them!