Question

$\underset{n\to \infty }{lim}\frac{(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdot \cdot (1-\frac{1}{n^2})}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}).(1-\frac{1}{n})}$ equals

• A. $\frac{2}{3}$
• B. $\infty$
• C. $\frac{5}{6}$
• D. $\frac{3}{5}$

Answer 1

The correct option is B $\infty$

Use the identity $a^2-b^2=(a+b)(a-b)$
thus
$\underset{n\to \infty }{lim}\frac{(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2}).\cdot .\cdot \cdot (1-\frac{1}{n^2})}{(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}).(1-\frac{1}{n})}$
$=\underset{n\to \infty }{lim}\frac{(1-\frac{1}{2})(1+\frac{1}{2})}{1-\left(\frac{1}{2}\right)}\frac{(1-\frac{1}{3})(1+\frac{1}{3})}{1-\left(\frac{1}{3}\right)}\frac{(1-\frac{1}{3})(1+\frac{1}{4})}{1-\left(\frac{1}{4}\right)}....\frac{(1-\frac{1}{n})(1+\frac{1}{n})}{1-\left(\frac{1}{n}\right)}$
$=\underset{n\to \infty }{lim}(1+\frac{1}{2})(1+\frac{1}{3})(1+\frac{1}{4})...(1+\frac{1}{n})$
$=\underset{n\to \infty }{lim}\frac{3}{2}\frac{4}{3}\frac{5}{4}....\frac{n+1}{n}$
$=\underset{n\to \infty }{lim}\frac{n+1}{2}$
Thus $\underset{n\to \infty }{lim}\frac{n+1}{2}$ also tends to infinity