Evaluating a limit with a factorial

[tmt1]

We are starting sequences, and in one of the examples we have this limit:

$$\lim_{{n}\to{\infty}} \frac{R^n}{n!}$$

We let $M$ equal a non-negative integer such that $ M \le R < M + 1$

I don't get the following step:

For $n > M$, we write $Rn/n!$ as a product of n factors:

$$\frac{R^n}{n!} = (\frac{R}{1} \frac{R}{2} ... \frac{R}{M}) (\frac{R}{M + 1}) (\frac{R}{M + 2}) ... (\frac{R}{n}) \le C(\frac{R}{n})$$

 

[Deveno]

Since $R < M+1$ all the factors $\dfrac{R}{M+k}$ for $k = 1,\dots,n-M-1$ are less than $1$, so we can replace them all by $1$.

That just leaves the factors $\dfrac{R^M}{M!}$ and $\dfrac{R}{n}$.

Thus $\dfrac{R^n}{n!} \leq \dfrac{R^{M+1}}{M!}\cdot \dfrac{1}{n}$

and we can let $C$ be the constant $\dfrac{R^{M+1}}{M!}$ (what this constant will be depends on what $R$ is).

The important thing about problems like this is that we're just looking for *some* bounding constant, it doesn't have to be a *tight* bound.

 

[tmt1]

Since $R < M+1$ all the factors $\dfrac{R}{M+k}$ for $k = 1,\dots,n-M-1$ are less than $1$, so we can replace them all by $1$.

That just leaves the factors $\dfrac{R^M}{M!}$ and $\dfrac{R}{n}$.

Thus $\dfrac{R^n}{n!} \leq \dfrac{R^{M+1}}{M!}\cdot \dfrac{1}{n}$

and we can let $C$ be the constant $\dfrac{R^{M+1}}{M!}$ (what this constant will be depends on what $R$ is).

The important thing about problems like this is that we're just looking for *some* bounding constant, it doesn't have to be a *tight* bound.

In that case, shouldn't it evaluate to $\frac{R^n}{n!} \le \frac{C}{n} $ where $C$ equals $\dfrac{R^{M+1}}{M!}$ instead of $\frac{R^n}{n!} \le \frac{CR}{n} $ ?

 

[Deveno]

In that case, shouldn't it evaluate to $\frac{R^n}{n!} \le \frac{C}{n} $ where $C$ equals $\dfrac{R^{M+1}}{M!}$ instead of $\frac{R^n}{n!} \le \frac{CR}{n} $ ?

Oops, it looks like I lumped too many $R$'s together (or I missed the $R$ in the numerator in the original problem).

So yes, we can take $C = \dfrac{R^M}{M!}$, and obtain $\dfrac{R^n}{n!} \leq C\dfrac{R}{n}$

It's not really going to matter, as soon as $n$ gets much larger than $CR$ (which is *still* a constant, so $n$ *will* get bigger eventually, because the reals are an Archimedean field), the ratio $\dfrac{CR}{n}$ is going to (eventually) get very small;

(for an official "epsilon" proof, you would take $n >> \dfrac{CR}{\epsilon}$ so that:

$\dfrac{R^n}{n!} \leq \dfrac{CR}{n} < (CR)\dfrac{\epsilon}{CR} = \epsilon$), which would show the limit is (rigorously) zero.

The thing to realize here, is that no matter how big a constant $R$ is, we can't exponentiate such an $R$ fast enough to overtake the factorial (factorials grow very fast), although for large values of $R$, the numerator may start growing faster "at first" (that why we need to be clever about picking $M$).

Sorry about the mix-up (can I blame my cell phone? No? How about Alzheimer's?).