Question

how do you determine if this series with factorials converges. \[\sum_{n=1}^{\infty} (2n)! \div (n!)^{2}\]

Answer 1

write the first couple of terms

Answer 2

it looks like you keep adding larger and larger numbers thus your series diverges

Answer 3

sorry the website has been glitchy I tried to answer earlier

Answer 4

\[\sum_{n=1}^{\infty} \frac{(2n)!}{(n!)^2}\]
Use the ratio test. If the ratio of successive terms approaches a finite number less than 1, then the series converges, and diverges otherwise. (Actually, if the limit turns out to be 1, you have to resort to another test, but the ratio test almost always works with factorials.)
\[a_n=\frac{(2n)!}{(n!)^2}\]
\[\begin{align*}\lim_{n\to\infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to\infty}\frac{(2(n+1))!}{((n+1)!)^2}\times\frac{(n!)^2}{(2n)!}\\\\
&=\lim_{n\to\infty}\frac{(2n+2)!}{((n+1)n!)^2}\times\frac{(n!)^2}{(2n)!}\\\\
&=\lim_{n\to\infty}\frac{(2n+2)(2n+1)(2n)!}{(n+1)^2(n!)^2}\times\frac{(n!)^2}{(2n)!}\\\\
&=\lim_{n\to\infty}\frac{(2n+2)(2n+1)}{(n+1)^2}\\\\
&=\lim_{n\to\infty}\frac{4n^2+\cdots}{n^2+\cdots}\end{align*}\]