# Math Help - A Sequential Limit

1. ## A Sequential Limit

Hi all, I was wondering if someone could help me show

$\lim_{n\rightarrow\infty}{2n \choose n}\left(\frac{1}{4}\right)^n=0$

I was able to do some algebra to show that this limit is

$\lim_{n\rightarrow\infty} \frac{1}{2}\frac{3}{4}\frac{5}{6}\dots \frac{2n-1}{2n}$

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

2. The 'infinite product' ...

$\displaystyle \prod_{k=1}^{\infty} (1+a_{k})$ (1)

... converges or diverges is converges or diverges the 'infinite series'...

$\displaystyle \sum_{k=1}^{\infty} a_{k}$ (2)

In Your case is $\displaystyle a_{k} = -\frac{1}{2k}$ so that the series (2) and the product (1) both diverge. The series tends to $-\infty$ so that the product tends to $0$...

Kind regards

$\chi$ $\sigma$

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

4. A 'good enough' tutorial can be found in...

Infinite product - Wikipedia, the free encyclopedia

Kind regards

$\chi$ $\sigma$