Does the series: $$\sum \frac{(n!)^2}{(2n)!}$$ converge or diverge?
I used the ratio test, and got an end result as $\lim_{n\to\infty}$ $\frac{n+1}{2}$ which would make it divergent but i know it's convergent. Am i using the right test?
[Math] Does the series $\sum$$(n!)^2/(2n)!$ diverge or converge
calculus
Best Answer
Your ratio test is incorrect. Note that $$\dfrac{\dbinom{2n}n}{\dbinom{2n+2}{n+1}} = \dfrac{(2n)!}{n! \cdot n!} \dfrac{(n+1)! (n+1)!}{(2n+2)!} = \dfrac{(n+1)^2}{2(n+1)(2n+1)} = \dfrac{n+1}{2(2n+1)}$$ Hence, $$\lim_{n \to \infty} \dfrac{\dbinom{2n}n}{\dbinom{2n+2}{n+1}} = \lim_{n \to \infty} \dfrac{n+1}{2(2n+1)} = \dfrac14 < 1$$
Alternatively, from Stirling, we have that the binomial coefficient goes as $$\dbinom{2n}n \sim \dfrac{4^n}{\sqrt{\pi n}}$$
You should be able to conclude from either whether the series converges or diverges from this.