Question

The value of $\underset{n\to \infty }{lim}\prod _{r=1}^n{\left(\frac{n^2+r^2}{n^2}\right)}^{\frac{1}{n}}$ is:

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

Answer 1

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

Let $P=\underset{n\to \infty }{lim}\prod _{r=1}^n{\left(\frac{n^2+r^2}{n^2}\right)}^{\frac{1}{n}}$
$=\underset{n\to \infty }{lim}{[\left(\frac{n^2+1^2}{n^2}\right)\left(\frac{n^2+2^2}{n^2}\right)\cdots \left(\frac{n^2+n^2}{n^2}\right)]}^{\frac{1}{n}}$
Taking $\ln$ on both sides, we have
$\ln P=\underset{n\to \infty }{lim}\frac{1}{n}[\ln (1+\frac{1^2}{n^2})+\ln (1+\frac{2^2}{n^2})+\cdots +\ln (1+\frac{n^2}{n^2})]\Rightarrow \ln P=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\ln (1+\frac{r^2}{n^2})=\underset{0}{\overset{1}{\int }}\ln (1+x^2)dx$
Using By parts, considering $\ln (1+x^2)$ as first function and $1$ as the second function, we have:
$\ln P={\left[x\ln (1+x^2)\right]}_0^1-\underset{0}{\overset{1}{\int }}x\left(\frac{2x}{1+x^2}\right)dx$
$=\ln 2-\underset{0}{\overset{1}{\int }}2(1-\frac{1}{1+x^2})dx=\ln 2-2{[x-{\tan }^{-1}x]}_0^1$
$=\ln 2-2(1-\frac{\pi }{4})=\ln 2+\frac{\pi -4}{2}\Rightarrow P=2e^{\left(\frac{\pi -4}{2}\right)}$