Question

Let $u_n=1+2+3+\dots +n$ and $v_n=\frac{u_2}{u_2-1}\cdot \frac{u_3}{u_3-1}\cdot \frac{u_4}{u_4-1}\cdots \frac{u_n}{u_n-1}$ where $n\ge 2$. Then find $\underset{n\to \infty }{\lim }{v}_n$.

Answer 1

Answer: 3

$u_n=1+2+3+\dots +n$
$=\frac{n(n+1)}{2}$
$\Rightarrow u_n-1=\frac{n^2+n-2}{2}=\frac{(n+2)(n-1)}{2}$
$\Rightarrow \frac{u_n}{u_n-1}=\left(\frac{n}{n-1}\right)\left(\frac{n+1}{n+2}\right)$
$\Rightarrow v_n=\left(\frac{2}{1}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \frac{5}{4}\dots \frac{n}{n-1}\right)$
$\left(\frac{3}{4}\cdot \frac{4}{5}\cdot \frac{5}{6}\dots \frac{n+1}{n+2}\right)$
$=\left(\frac{n}{1}\right)\left(\frac{3}{n+2}\right)$
$\Rightarrow \underset{n\to \infty }{\lim }{v}_n=3\underset{n\to \infty }{\lim }\frac{n}{n+2}$
$=3\underset{n\to \infty }{\lim }\frac{1}{1+\frac{2}{n}}=3$