Question

Let $u_{n}=1+2+3+\ldots +n$ 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 $n \geq 2$. Then find $\displaystyle\lim _{n \rightarrow \infty} v_{n}$.

Answer 1

$u_{n} =1+2+3+\ldots +n$
$=\frac{n(n+1)}{2}$
$\Rightarrow u_{n}-1=\frac{n^{2}+n-2}{2}=\frac{(n+2)(n-1)}{2}$
$\Rightarrow \frac{u_{n}}{u_{n}-1}=\left(\frac{n}{n-1}\right)\left(\frac{n+1}{n+2}\right)$
$\Rightarrow v_{n}=\left(\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \ldots \frac{n}{n-1}\right)$
$\left(\frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \ldots \frac{n+1}{n+2}\right)$
$=\left(\frac{n}{1}\right)\left(\frac{3}{n+2}\right)$
$\Rightarrow \displaystyle\lim _{n \rightarrow \infty} v_{n}=3 \displaystyle\lim _{n \rightarrow \infty} \frac{n}{n+2}$
$=3 \displaystyle\lim _{n \rightarrow \infty} \frac{1}{1+\frac{2}{n}}=3 $