# Use Ratio Test to find the convergence of the following series?

## sum_(n=0)^oo ((2n)!)/((3^n)(n!)^2

Apr 9, 2018

The series is divergent, because the limit of this ratio is >1

${\lim}_{n \to \infty} {a}_{n + 1} / {a}_{n} = {\lim}_{n \to \infty} \frac{4 \left(n + \frac{1}{2}\right)}{3 \left(n + 1\right)} = \frac{4}{3} > 1$

#### Explanation:

Let ${a}_{n}$ be the n-th term of this series:
a_n=((2n)!)/(3^n(n!)^2)

Then
a_(n+1)=((2(n+1))!)/(3^(n+1)((n+1)!)^2)

=((2n+2)!)/(3*3^n((n+1)!)^2)

=((2n)!(2n+1)(2n+2))/(3*3^n(n!)^2(n+1)^2)

=((2n)!)/(3^n(n!)^2)*((2n+1)(2n+2))/(3(n+1)^2)

$= {a}_{n} \cdot \frac{\left(2 n + 1\right) 2 \left(n + 1\right)}{3 {\left(n + 1\right)}^{2}}$

${a}_{n + 1} = {a}_{n} \cdot \frac{2 \left(2 n + 1\right)}{3 \left(n + 1\right)}$

${a}_{n + 1} / {a}_{n} = \frac{4 \left(n + \frac{1}{2}\right)}{3 \left(n + 1\right)}$

Taking limit of this ratio

${\lim}_{n \to \infty} {a}_{n + 1} / {a}_{n} = {\lim}_{n \to \infty} \frac{4 \left(n + \frac{1}{2}\right)}{3 \left(n + 1\right)} = \frac{4}{3} > 1$

So the series is divergent.