Question

The value of ${lim}_{n\to \infty }{\left\{\frac{(n^3+1)(n^3+2^3)(n^3+3^3)........(n^3+n^3)}{n^3}\right\}}^{\frac{1}{n}}$ is equal to $\alpha .e^{\beta \int \frac{x^3}{1+x^3}dx}$, where $\alpha ϵN,\beta ϵR$,then find $\alpha -\beta$.___

Answer 1

$y={lim}_{n\to \infty }{\left\{\frac{(n^3+1)(n^3+2^3)(n^3+3^3)........(n^3+n^3)}{n^3}\right\}}^{\frac{1}{n}}$
Taking log on both sides, we get
$logy={lim}_{n\to \infty }\frac{1}{n}\left\{{\sum }_{r=1}^{n}log\right(n^3+r^3)-log{n}^3\}\Rightarrow logy={lim}_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^{n}log[1+{\left(\frac{r}{n}\right)}^3]\Rightarrow logy={\int }_0^{1}log(1+x^3)dx\Rightarrow logy=\left[log\right(1+x^3).x{]}_0^1-{\int }_0^1\frac{3x^2}{1+x^3}xdx\Rightarrow logy=log2-3{\int }_0^1\frac{x^2}{1+x^3}xdx\Rightarrow y=2.e^{-3{\int }_0^1}\frac{x^3}{1+x^3}\therefore \alpha -\beta =2-(-3)=5$