Does the series converge absolute, converge conditionally, or diverge.$$\sum_{n=1}^{\infty} \frac{{(-1)}^n {2n \choose n}}{3^n}$$
I am confuse because doesnt $$\lim_{n \to \infty} \frac{{2n \choose n}}{3^n}=0$$? Why doesnt this converge condiitonally at least and maybe absolutely? Key says it diverges.
The central binomial coefficient satisfies $\dbinom{2n}{n} \sim \dfrac{4^n}{\sqrt{\pi n}}$ as $n \to \infty$.
Hence, $\dfrac{\binom{2n}{n}}{3^n} \sim \dfrac{(4/3)^n}{\sqrt{\pi n}}$ as $n \to \infty$, and thus, $\displaystyle\lim_{n \to \infty}\dfrac{\binom{2n}{n}}{3^n} = \infty$. So the series diverges.