Ratio Test — Question 4

*Question 4 Analyze $\displaystyle\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^2\,5^n}$. Compute the consecutive-term ratio carefully, apply the Ratio Test, and state the convergence classification.

*Question 4 – Solution

Let $a_n=(2n)!/[(n!)^2 5^n]$.

$$\begin{align*} \frac{a_{n+1}}{a_n}&=\frac{(2n+2)!}{((n+1)!)^2 5^{n+1}}\frac{(n!)^2 5^n}{(2n)!}\\&=\frac{(2n+2)(2n+1)}{5(n+1)^2}=\frac{2(2n+1)}{5(n+1)}. \end{align*}$$

$\displaystyle L=\lim_{n\to\infty}\frac{2(2n+1)}{5(n+1)}=\frac45<1$.

The positive-term series converges. The factorial expression is $\binom{2n}{n}$, but no approximation is needed.