1
$\begingroup$

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!

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ Not all the limits are $0$. For example the last one isn't. $\endgroup$ – user258700 Nov 26 '15 at 16:50
  • $\begingroup$ Yes, that's what I felt while solving this problem. In fact, the last limit is $ - \infty $. However, I wasn't able to think of any other way to proceed. $\endgroup$ – Kaustav Sen Nov 26 '15 at 16:52
  • $\begingroup$ Are you sure that you have typed in everything correctly? $\endgroup$ – user258700 Nov 26 '15 at 16:52
  • $\begingroup$ Yes, I just cross-checked it with the question given in the book $\endgroup$ – Kaustav Sen Nov 26 '15 at 16:54
  • 1
    $\begingroup$ What number is m? Natural or any real mumber? $\endgroup$ – kmitov Nov 26 '15 at 16:56
2
$\begingroup$

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 $\lim_n p_2\cdots p_n = 0$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

If you know the gamma function:

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ And what hapens then? $\endgroup$ – kmitov Nov 26 '15 at 21:18
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why the downvote? $\endgroup$ – copper.hat Nov 26 '15 at 16:59
  • $\begingroup$ I wanted to ask the same. $\endgroup$ – kmitov Nov 26 '15 at 16:59
  • 1
    $\begingroup$ Welcome to the club of victims of downvoters ! $\endgroup$ – Claude Leibovici Nov 26 '15 at 17:22
  • $\begingroup$ What do you meen? $\endgroup$ – kmitov Nov 26 '15 at 21:17
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.