Limit of an exotic sequence Let $a_m = \frac{ m! \cdot m!}{(m-k)! \cdot (m+k)!} $. Is there a way to show that 
$$ \lim ( a_m)^m = e^{-k^2} $$
???
I was trying to remove the factorials as follows: 
$$ m! = m(m-1)(m-2)...(m-(k+1))(m-k)! $$
$$ (m+k)! = (m+k-1)(m+k-2)...(m + k -(k-1)) m! $$
And so after cancelation, we would have 
$$ a_m = \frac{ m(m-1)(m-2)...(m-(k+1)) }{(m+k-1)(m+k-2)...(m + k -(k-1))} $$
But this still looks hard to manage.
 A: After cancelling common factors in the numerator and denominator, $a_m$ becomes
$$a_m=\frac{m(m-1)\cdots(m-(k-1))}{(m+1)(m+2)\cdots(m+k)},$$
and dividing both numerator and denominator by $m^k$ then gives
$$
a_m=\frac{ (1-\frac{1}{m})(1-\frac{2}{m})\cdots(1-\frac{k-1}{m}) }{(1+\frac{1}{m})(1+\frac{2}{m})\cdots(1+\frac{k}{m})}.
$$.
Therefore,
$$a_m^m=\frac{ (1-\frac{1}{m})^m(1-\frac{2}{m})^m\cdots(1-\frac{k-1}{m}) ^m}{(1+\frac{1}{m})^m(1+\frac{2}{m})^m\cdots(1+\frac{k}{m})^m}. \ \ \ (*)$$
Now use the fact that, for any fixed $r$,
$$
\lim_{m\to\infty} (1+\frac{r}{m})^m=e^r.
$$
In the limit, $(*)$ then becomes
$$
\frac{ e^{-1}\cdots e^{-(k-1)}}{e e^2\cdots e^k}=\frac{e^{-k(k-1)/2}}{e^{k(k+1)/2}}=e^{-k^2}.$$
A: Another way to solve this is to L'Hospital's Rule. In fact, let $y_m=\ln((a_m)^m)$. Then
\begin{eqnarray}
y_m&=&m\ln a_m\\
&=&m\sum_{i=0}^{k-1}[\ln(m-i)-\ln(m+i+1)]\\
&=&\sum_{i=0}^{k-1}\frac{\ln(1-\frac im)-\ln(m+\frac{i+1}m)}{\frac 1m}.
\end{eqnarray}
So
\begin{eqnarray}
\lim_{m\to\infty}y_m
&=&\lim_{m\to\infty}\sum_{i=0}^{k-1}\frac{\ln(1-\frac im)-\ln(m+\frac{i+1}m)}{\frac 1m}\\
&=&\lim_{x\to 0}\sum_{i=0}^{k-1}\frac{\ln(1-ix)-\ln(1+(i+1)x)}{x}\\
&=&\lim_{x\to 0}\sum_{i=0}^{k-1}\left(\frac{-i}{1-ix}-\frac{i+1}{1+(i+1)x}\right)\\
&=&-\sum_{i=0}^{k-1}(2i+1)\\
&=&-k^2
\end{eqnarray}
and hence
$$ \lim_{m\to\infty}a_m^m=e^{-k^2}. $$
