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Calculus Level 4

$\begin{aligned} \lambda&=\int_{0}^{1}\frac{dx}{1+x^{3}} \\ p&=\lim_{n\to \infty} \left[{\frac{\prod_{r=1}^{n}(n^{3}+r^{3})}{n^{3n}}}\right] ^{\frac{1}{n}}\end{aligned}$ If $\lambda$ and $p$ are given as above, find $\ln(p)$ in terms of $\lambda$ .

\ln(4)-3+3\lambda

\ln(2)-3+3\lambda

2\ln(2)-\lambda

\ln(4)-1+3\lambda

1 solution

Rishabh Jain
Jun 6, 2016

$\small{p=\displaystyle\lim_{n\to \infty} \left[{\dfrac{\displaystyle\prod_{r=1}^{n}(n^{3}+r^{3})}{(n^{3})^{\large ^n}}}\right] ^{\frac{1}{n}}= \displaystyle\lim_{n\to \infty} \left[\displaystyle\prod_{r=1}^{n}\left(1+\dfrac{r^{3}}{n^3}\right)\right] ^{\frac{1}{n}}}$

$\ln p= \displaystyle\lim_{n\to \infty} \dfrac 1n \left[\displaystyle\sum_{r=1}^{n}\left(1+\dfrac{r^{3}}{n^3}\right)\right]$

By Reimann Sums , this is:

$\large \displaystyle\int_0^1\ln(1+x^3)\mathrm{d}x$

Use IBP :

$=\left[x\ln(1+x^3)\right]_0^1-3\displaystyle\int_0^1\underbrace{\dfrac{x^3}{1+x^3}}_{1-\frac{1}{1+x^3}}\mathrm{d}x$

$\large =\ln2-3(1-\lambda)$ $\Large =\boxed{\ln2-3+3\lambda}$

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