Does $\sum\frac{4^n(n!)^2}{(2n)!}$ converge? The ratio test is inconclusive. I know that $\sum\frac{2^n(n!)^2}{(2n)!}$ converges, if that's of any help.
$\endgroup$ 17 Answers
$\begingroup$One of the terms of the expansion of $(1+1)^{2n}$ is ${2n \choose n}$ so $\frac{4^n}{{2n \choose n}}\geq 1$ which means the sum diverges.
$\endgroup$ $\begingroup$In order for a series $\sum a_n$ to converge, we must have $\lim_{n\to\infty} a_n = 0$.
What is the $\lim_{n\to\infty} \frac{4^n(n!)^2}{(2n)!}$?
$\endgroup$ $\begingroup$Making the problem more general, consider $$u_n=\frac{a^n(n!)^2 }{(2 n)!}\implies \frac{u_{n+1}}{u_n}=a\frac{ (n+1)}{2(2 n+1)}$$ When $n\to \infty$ the limit is $\frac a 4$ and then you have convergence if $|a|<4$.
$\endgroup$ $\begingroup$it is divergent however it is borel summable.
$\endgroup$ $\begingroup$Use the limit comparison theorem combined with Stirling's formula.
$\endgroup$ $\begingroup$Try the ratio test. This series diverges.
$\endgroup$ $\begingroup$Try
$$ a_n = \frac{4^n (n!)^2}{(2n)!} $$$$ \frac{a_{n+1}}{a_{n}} = 4 \frac{(n+1)^2}{(2n+1)(2n+2)} > 1 $$
conclude that $a_n$ is monotonically increasing and also $a_n > 0$ so$$ \lim \limits_{n \to \infty} a_n \ne 0 $$So$\sum a_n $ diverge.
$\endgroup$
