Ratio Test — Question 4

Question 4 – Solution

Step 1: Name the term.

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

Step 2: Expand only the new factorial factors.

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

Step 3: Take the limit.

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

Conclusion.

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