439 views

Lower bounds for ${{2n} \choose {n}}$ [duplicate]

What are common lower bounds for ${{2n} \choose {n}}$? Edit: I made a mistake in my original question. It doesn't change my question but there is no reason for me to include the mistake.
221 views

9k views

Calculating the limit of $[(2n)!/(n!)^2]^{1/n}$ as $n$ tends to $\infty$

Analysis textbook by Shanti Narayan, is asking to prove the limit $\lim {\left({\dfrac{(2n)!}{(n!)^2}}\right)}^{1/n} \to \frac{1}{4}$ as $n \to \infty$. I tried but was unable to find the solution. ...
679 views

Inequality $\binom{2n}{n}\leq 4^n$

I would like to prove the following inequality, for $n=0,1,2,...$, $$\binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
384 views

Approximate solution: factorial and exponentials

If z= $\dbinom{200}{100}/(4^{100})$, what is the value of z? The options are: a. $z<1/3$ b. $1/3<z<1/2$ c. $1/2<z<2/3$ d. $2/3<z<1$ How should I go about solving these type ...
Computing $\lim\limits_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum\limits_{k=1}^{n} \binom{2n-1}{n-k}\frac{ 1}{(2k-1)^2+\pi^2}$
What tools would you recommend me for computing the limit below? $$\lim_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum_{k=1}^{n}\frac\binom{2n-1}{n-k}}{(2k-1)^2+\pi^2$$ As soon as any ...