Question

Evaluate $\underset{n\to \infty }{lim}{[(1+\frac{1}{n^2})(1+\frac{2^2}{n^2})(1+\frac{3^2}{n^2})......(1+\frac{n^2}{n^2})]}^{1/n}$

• A. $2e^{\left(\frac{\pi -4}{2}\right)}$
• B. $2e^{\left(\frac{\pi -2}{2}\right)}$
  
• C. $2e^{\left(\frac{\pi -4}{4}\right)}$
• D. $2e^{\left(\frac{\pi -4}{3}\right)}$

Answer 1

The correct option is A $2e^{\left(\frac{\pi -4}{2}\right)}$

Let $S=\frac{lim}{n\to \infty }{[(1+\frac{1}{n^2})(1+\frac{2^2}{n^2})...(1+\frac{n^2}{n^2})]}^{\frac{1}{n}}$
$\Rightarrow logS=\frac{lim}{n\to \infty }\frac{1}{n}\sum _{r=1}^{n}log(1+\frac{r^2}{n^2})$
$={\int }_0^{1}log(1+x^2)dx.={\left(xlog(1+x^2)\right)}_0^1-{\int }_0^1\frac{{2x}^2}{1+x^2}dx$
$\Rightarrow log\left(\frac{S}{2}\right)=2[{\int }_0^1\frac{dx}{1+x^2}-{\int }_0^{1}dx]=2[\frac{\pi }{4}-1]=\frac{\pi -4}{2}$
$\Rightarrow S=2e^{\left(\frac{\pi -4}{2}\right)}$