Evaluating the limit $\mathop {\lim }\limits_{n \to \infty } \frac{{m(m - 1)....(m - n + 1)}}{{(n - 1)!}}{x^n}$

Question

Evaluate: $$\mathop {\lim }\limits_{n \to \infty } \frac{{m(m - 1)....(m - n + 1)}}{{(n - 1)!}}{x^n}$$ where $x \in (-1,\ 1)$

I tried solving this as follows: $$\mathop {\lim }\limits_{n \to \infty } \frac{{mx}}{{n - 1}}\frac{{(m - 1)x}}{{n - 2}}.....\frac{{(m - n + 1)x}}{1}$$ $$=\{ \mathop {\lim }\limits_{n \to \infty } \frac{{mx}}{{n - 1}}\} \{ \mathop {\lim }\limits_{n \to \infty } \frac{{(m - 1)x}}{{n - 2}}\} .....\{ \mathop {\lim }\limits_{n \to \infty } \frac{{(m - n + 1)x}}{1}\} $$

Now as the individual limit $\ \mathop {\lim }\limits_{n \to \infty } \frac{{mx}}{{n - 1}}$ evaluates to $0$ can we say that the limit as whole tends to $0$?

I personally feel that this reasoning seems incorrect. Can someone please suggest an alternate way of tackling this problem.

Thanks!

Answer 1

Note that let $p_n = {m-n+1 \over n-1}x$ and $\lim_n p_n = -x$.

Take $|x|< \beta < 1$, then for some $N$ we have $|p_n| < \beta$ for all $n \ge N$.

Then $|p_2\cdots p_n | \le |p_2 \cdots p_{N}| \beta^{n-N} $, hence $\lim_n p_2\cdots p_n = 0$.

Answer 2

If you know the gamma function:

$$\lim_{n\to\infty}\left(\frac{{m(m - 1)....(m - n + 1)}}{{(n - 1)!}}\cdot x^n\right)=\lim_{n\to\infty}\left(\frac{\prod_{k=1}^{n}\left(m-k+1\right)}{{(n - 1)!}}\cdot x^n\right)=$$ $$\lim_{n\to\infty}\left(\frac{\frac{m!}{(m-n)!}}{{(n - 1)!}}\cdot x^n\right)=\lim_{n\to\infty}\left(\frac{\frac{m!}{(m-n)!}}{{(n - 1)!}}\cdot x^n\right)=$$ $$\lim_{n\to\infty}\left(\frac{nm!}{n!(m-n)!}\cdot x^n\right)=\lim_{n\to\infty}\left(\frac{x^nm!}{\Gamma(n)\Gamma(m-n+1)}\right)=$$ $$m!\lim_{n\to\infty}\left(\frac{x^n}{\Gamma(n)\Gamma(m-n+1)}\right)=m!\lim_{n\to\infty}\left(\frac{x^n}{\Gamma(n)(m-n)!}\right)$$

Answer 3

If $m$ is a positive integer the limit is evidently zero.

Answer 4

Hint: Basically, you are asking us to evaluate $\displaystyle\lim_{n\to\infty}n~{m\choose n}~x^n$ for $m\not\in\mathbb Z$ and $|x|<1$. The obvious choice would be Stirling's approximation.