Question

Show that the product
$\frac{2}{1}.\frac{2}{3}.\frac{4}{3}.\frac{4}{5}.\frac{6}{5}....\frac{2n-2}{2n-3}.\frac{2n-2}{2n-1}.\frac{2n}{2n-1}$
is finite when $n$ is infinite.

Answer 1

Let the general term of series $a_n=\frac{2n}{2n-1}\cdot \frac{2n}{2n+1}=\frac{4n^2}{4n^2-1}$

then ${\Pi }_{n=1}^{\infty }{a}_n=\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3}\cdot \frac{4}{5}\cdot \frac{6}{5}\cdot \frac{6}{7}...$

$a_{n+1}=\frac{4(n+1{)}^2}{\left(4\right(n+1{)}^2-1)}{a}_n=\frac{4n^2}{4n^2-1}$

Applying ratio test,

$\underset{n\to \infty }{lim}\frac{a_{n+1}}{a_n}=\frac{4(n+1{)}^2}{\left(4\right(n+1{)}^2-1)}\times \frac{4n^2-1}{4n^2}=\underset{n\to \infty }{lim}\frac{4(1+1/n{)}^2(4-1/n^2)}{[4{(1+\frac{1}{n})}^2-\frac{1}{n^2}]4}=\underset{n\to \infty }{lim}\frac{4\times 4}{4\times 4}=1$

which is a finite number