Question

The value of $\underset{n\to \infty }{lim}\prod _{r=2}^n\frac{r^3+1}{r^3-1}$ is

(Here, $\prod$ stands for the product.)

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{3}{2}$

We have,
$\underset{n\to \infty }{lim}\prod _{r=2}^n\frac{r^3+1}{r^3-1}$
$=\underset{n\to \infty }{lim}\{\left(\frac{2^3+1}{2^3-1}\right)\left(\frac{3^3+1}{3^3-1}\right)\left(\frac{4^3+1}{4^3-1}\right)...\left(\frac{n^3+1}{n^3-1}\right)\}$

$=\underset{n\to \infty }{lim}\{\left[\frac{2+1}{2-1}\frac{2^2-2+1}{2^2+2+1}\right]\left[\frac{3+1}{3-1}\frac{3^2-3+1}{3^2+3+1}\right]\left[\frac{4+1}{4-1}\frac{4^2-4+1}{4^2+4+1}\right]...\left[\frac{n+1}{n-1}\frac{n^2-n+1}{n^2+n+1}\right]\}$

$=\underset{n\to \infty }{lim}\left[\frac{\mathrm{3.4.5...}(n+1)}{\mathrm{1.2.3...}(n-1)}\right]\left[\frac{\mathrm{3.7.13...}(n^2-n+1)}{\mathrm{7.13.21...}(n^2+n+1)}\right]$

$=\underset{n\to \infty }{lim}\left[\frac{n(n+1)}{1.2}\frac{3}{n^2+n+1}\right]$
$=\frac{3}{2}\underset{n\to \infty }{lim}\frac{n^2+n}{n^2+n+1}$
$=\frac{3}{2}\underset{n\to \infty }{lim}\frac{1+\frac{1}{n}}{1+\frac{1}{n}+\frac{1}{n^2}}$
$=\frac{3}{2}$