Question

Let $\lambda =\underset{0}{\overset{1}{\int }}\frac{dx}{1+x^3}$ and $p=\underset{n\to \infty }{lim}{\left(\frac{\prod _{r=1}^n(n^3+r^3)}{n^{3n}}\right)}^{1/n}.$ Then the value of $\ln p$ is equal to

• A. $\ln 2-1+3\lambda$
• B. $\ln 2-3+3\lambda$
• C. $2\ln 2+3\lambda$
• D. $\ln 4-3+3\lambda$

Answer 1

The correct option is B $\ln 2-3+3\lambda$

$p=\underset{n\to \infty }{lim}{\left(\frac{\prod _{r=1}^n(n^3+r^3)}{n^{3n}}\right)}^{1/n}$
Taking $\ln$ on both sides, we have
$\ln p=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\ln (1+{\left(\frac{r}{n}\right)}^3)=\underset{0}{\overset{1}{\int }}\ln (1+x^3)dx$
Using By-parts, with $\ln (1+x^3)$ as the first function and $1$ as the second function, we have
$\ln p={\left[x\ln \right(1+x^3\left)\right]}_0^1-\underset{0}{\overset{1}{\int }}\frac{x\cdot 3x^{2}dx}{1+x^3}=\ln 2-3\underset{0}{\overset{1}{\int }}(1-\frac{1}{1+x^3})dx=\ln 2-3+3\lambda$