
A Sequential Limit
Hi all, I was wondering if someone could help me show
$\displaystyle \lim_{n\rightarrow\infty}{2n \choose n}\left(\frac{1}{4}\right)^n=0$
I was able to do some algebra to show that this limit is
$\displaystyle \lim_{n\rightarrow\infty} \frac{1}{2}\frac{3}{4}\frac{5}{6}\dots \frac{2n1}{2n}$
which must converge since the sequence is decreasing and bounded by 0. But I haven't been able to figure out how to show the limit is actually 0 using the epsilon/delta definition of convergence. Is there an alternative way that I might prove the limit is 0, or can someone give me a hint with regard to the epsilon/delta definition?
Thanks

The 'infinite product' ...
$\displaystyle \displaystyle \prod_{k=1}^{\infty} (1+a_{k})$ (1)
... converges or diverges is converges or diverges the 'infinite series'...
$\displaystyle \displaystyle \sum_{k=1}^{\infty} a_{k}$ (2)
In Your case is $\displaystyle \displaystyle a_{k} = \frac{1}{2k}$ so that the series (2) and the product (1) both diverge. The series tends to $\displaystyle \infty$ so that the product tends to $\displaystyle 0$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$

Thanks for the help chisigma, but I'm a little confused still. Why does the sum diverging to $\displaystyle \infty$ imply that the product goes to $\displaystyle 0$? I assume that it has something to do with the logarithmic relationship between sums and products and the fact that $\displaystyle \lim_{x\rightarrow \infty} log(x)=0$, but I'm not sure how you derive that theorem between the product and the series.

A 'good enough' tutorial can be found in...
Infinite product  Wikipedia, the free encyclopedia
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$