The limit $\lim\limits_{n\to\infty}\frac{x^n}{n!}$

Question

Evaluate $$\lim_{n\to\infty}\frac{x^n}{n!}$$ where $x\in\mathbb R$

This is a common limit and has been asked and answered many times here on this site. However, I present another approach with Stolz-Cesaro($x\notin[-1,1]$. When $x\in[-1,1]$, the limit is trivially zero):

$$L=\lim_{n\to\infty}\frac{x^n}{n!}=\lim_{n\to\infty}\frac{x^{n+1}-x^n}{(n+1)!-n!}$$

$$L=\lim_{n\to\infty}\frac{x^n}{n!}\frac{(x-1)}{n}=L\frac{(x-1)}{n}$$

The above equation holds for all $x$ in domain only when $L=0$. Hence the limit is just $0$.

I have never seen Stolz-Cesaro being used this way to solve a limit, hence the lack of surety. It would be great to get this verified!

Answer 1

You have not proved that the limit exists, what Stolz-Cesaro criterion says is the following:

Let $\{a_n\}$ and $\{b_n\}$ be sequences, with $\{b_n\}$ monotone increasing and divergent. Then, if $$\lim_{n\rightarrow \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lambda,\ \lambda \in \mathbb{R} \Rightarrow \lim_{n\rightarrow \infty}\frac{a_n}{b_n}=\lambda$$ In order to use it you would have to calculate the limit after the second $=$ and then the criterion would guarantee that the limit you are actually trying to calculate is equal.

The most immediate way to see why it is zero is to notice that you are trying to calculate the limit of the general term of a convergent series (in this case the expansion of $e^x$), so it has to be zero.

Answer 2

Alternative (much less elegant) approach than the one taken in the answer of José Antonio.

Given:
$~x~$ is some fixed (i.e. constant) real number.

It is sufficient to prove that
$~\forall ~\epsilon > 0, ~\exists ~M \in \Bbb{Z^+},~$ such that for all $~\displaystyle n \geq 2M, ~\left| ~\frac{x^n}{n!} ~\right| < \epsilon.$

---

For any fixed $~M \in \Bbb{Z^+},~$
let $~f(k)~$ denote $(1 + 2M - k) \times k ~: 1 \leq k \leq M.$

Then, $~f'(k) = (1 + 2M) - 2k > 0 ~: ~1 \leq k \leq M.~$
Therefore, $~f(k)~$ is strictly increasing on the interval $~1 \leq k \leq M.$
Therefore, $(1 + 2M - k)k = f(k) \geq f(1) = 2M.$

Then

$$\left| ~\frac{x^{2M}}{(2M)!} ~\right| = \prod_{k=1}^M \frac{x^2}{f(k)} \leq \left[ ~\frac{x^2}{2M} ~\right]^M = \left[ ~\frac{x^2}{M} ~\right]^M \times \frac{1}{2^M}. \tag1 $$

Given a fixed $~x \in \Bbb{R},~$ and given a fixed $~\epsilon > 0,~$ choose $~M \in \Bbb{Z^+},~$ such that:

• $~x^2 < M.$
• $~\dfrac{1}{2^M} < \epsilon.$

For example, using $~\lceil r\rceil~$ to denote the ceiling of $~r,~$ you can set

$$M = 1 + \max\left\{ ~\left\lceil ~x^2 ~\right\rceil, ~~\left\lceil ~\frac{\ln(1/\epsilon)}{\ln(2)} ~\right\rceil ~\right\}. \tag2 $$

With $~M~$ chosen by the constraint in (2) above, the RHS of (1) above is strictly $~\displaystyle < 1 \times \frac{1}{2^M} < \epsilon.$

Further, since $~M \in \Bbb{Z^+},~$ and $~x^2 < M,~$ you have that $~|x| < M < 2M.$

The problem can now be completed by induction.

Let $~N~$ be any positive integer such that :

• $~|x| < N.$

• $\displaystyle \left| ~\frac{x^{N}}{N!} ~\right| < \epsilon.$

Then:

• $\displaystyle |x| < (N+1).$

• $\displaystyle \left| ~\frac{x^{N+1}}{(N+1)!} ~\right| = \left| ~\frac{x}{N+1} ~\right| \times \left| ~\frac{x^{N}}{N!} ~\right| < \epsilon.$

So, if the two inequalities above hold for $~N,~$ they also hold for $(N+1).$

Therefore, with $~M~$ set by the constraints in (2) above, for all $~n \in \Bbb{Z^+}~$ such that $~n \geq 2M,~$ the two inequalities will hold for $~n.$