$\lim_{n\to\infty}\frac{(2n-1)!}{3^n(n!)^2}$
How can I associate limit problem with series? And how can i find limits from series? Can anyone help?
$\lim_{n\to\infty}\frac{(2n-1)!}{3^n(n!)^2}$
How can I associate limit problem with series? And how can i find limits from series? Can anyone help?
Hint: Let $a_n=\dfrac{(2n-1)!}{3^n(n!)^2}$.
It is useful to look at the ratio $\dfrac{a_{n+1}}{a_n}$ for large $n$.
by ratio rule:
$\dfrac{(2(n+1)-1)!}{3^{n+1}((n+1)!)^2}\cdot\dfrac{3^n(n!)^2}{(2n-1)!}= \dfrac{4n^2+2n}{3n^2+6n+3} \rightarrow \dfrac{4}{3}$
thus the series doesnot converge as the quotient and thus limsup is bigger than 1