More fun in 2016 part 2

$\displaystyle\lim _{ n\rightarrow \infty }{ \displaystyle\int _{ 0 }^{ 1 }{ \cdots \displaystyle\int _{ 0 }^{ 1 }{ \sqrt [ n ]{ \displaystyle \left( \prod _{ k=1 }^{ n }{ { x }_{ k } } \right )^{ 2016 }}d{ x }_{ 1 }d{ x }_{ 2 }\cdots d{ x }_{ { n } } } } }$

$\text{ Let } \ L = \displaystyle \lim_{n \to \infty}{\sqrt[n]{ \left( \prod_{k=1}^{n}{x_k} \right )^{2016}}}$

$\implies \ln{L} = \displaystyle \lim_{n \to \infty}{ \left( \dfrac{2016}{n}\sum_{k=1}^{n}{\ln{{x_k}}} \right )}$ $\displaystyle = 2016 \int_0^1 {\ln{(x_n)}} dx_n$

$\implies \ln{L} = 2016x_n(\ln(x_n)-1)|_0^1 = -2016 \\ \therefore L = e^{-2016}$

So $a+b = 2017$ .

$\displaystyle\int_0^1...\int_0^1\bigg(\prod_{k=1}^\infty x_k\bigg)^\frac{2016}{n}dx_1...dx_n=\int_0^1x_1^\frac{2016}{n}dx_1\times...\times\int_0^1x_n^\frac{2016}{n}dx_n=$

$\displaystyle\underbrace{\frac1{1+\frac{2016}{n}}\times...\times\frac1{1+\frac{2016}{n}}}_\text{n times}=\frac1{(1+\frac{2016}{n})^n}$

$\displaystyle\lim_{n\to\infty}\frac1{(1+\frac{2016}{n})^n}=\frac1{e^{2016}}$