Question

Let $S_n=1+2+3+...+n$ and $P_n=\frac{S_2}{S_2-1}\cdot \frac{S_3}{S_3-1}\cdot \frac{S_4}{S_4-1}\cdot \cdot \cdot \frac{S_n}{S_n-1}$ where $n\in N,(n\ge 2).$ Then $\underset{n\to \infty }{lim}{P}_n=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Given, $S_n=1+2+3+...+n$
$S_n=\frac{n(n+1)}{2}$
So $S_n-1=\frac{(n+2)(n-1)}{2}$
$\therefore \frac{S_n}{S_n-1}=\frac{n(n+1)}{2}\cdot \frac{2}{(n+2)(n-1)}$
$=\left(\frac{n}{n-1}\right)\left(\frac{n+1}{n+2}\right)$
Given, $P_n=\frac{S_2}{S_2-1}\cdot \frac{S_3}{S_3-1}\cdot \frac{S_4}{S_4-1}\cdot \cdot \cdot \frac{S_n}{S_n-1}$
$\therefore P_n=(\frac{2}{1}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \frac{5}{4}\cdot \cdot \cdot \frac{n}{n-1})(\frac{3}{4}\cdot \frac{4}{5}\cdot \frac{5}{6}\cdot \cdot \cdot \frac{n+1}{n+2})$
$\Rightarrow P_n=\left(\frac{n}{1}\right)\left(\frac{3}{n+2}\right)$
$\therefore \underset{n\to \infty }{lim}{P}_n=\underset{n\to \infty }{lim}\frac{3n}{n+2}$

$=\underset{n\to \infty }{lim}\frac{3n}{n(1+\frac{2}{n})}=3$