Limits Or Not..

$\textbf{Solution}$

$\beta$ = $\displaystyle\lim_{n\rightarrow \infty} \displaystyle \prod _{ r=1 }^{ n }{ { \left( 1+\dfrac { r }{ n } \right) }^{ \dfrac { 1 }{ r } } }$

Taking Logarithm on Both Sides we get,

$\ln { \beta }$ = $\displaystyle\lim_{n\rightarrow \infty} \sum _{ r=1 }^{ n }{ \dfrac{1}{r}{ \ln\left( 1+\dfrac { r }{ n } \right) } }$

Multiplying And Dividing By $\textbf{n}$ ,We get

$\ln { \beta }$ = $\displaystyle\lim_{n\rightarrow \infty} \sum _{ r=1 }^{ n }{ \dfrac{1}{n} \dfrac{n}{r}{ \ln \left( 1+\dfrac { r }{ n } \right) } }$

Which Can be Rewritten as,

$\ln { \beta }$ = $\displaystyle\lim_{n\rightarrow \infty} \sum _{ r=1 }^{ n }{ \dfrac{1}{n} \dfrac{1}{(\dfrac{r}{n})}{ \ln \left( 1+\dfrac { r }{ n } \right) } }$

Now It can Be Written in Form of Integration As Using $\textbf{Riemann sums}$ ,

$\ln { \beta }$ = $\displaystyle\int_{0 }^{1}{\dfrac{\ln(1+x)}{x}} \mathrm{d}x$

Using Series Expansion Since $|x| \leq 1$ , We Get,

$\ln { \beta }$ = $\displaystyle\int_{0 }^{1}{\dfrac{\displaystyle \sum _{ n=1 }^{ \infty } \dfrac { (-1)^{ n+1 } }{ n } x^{ n }}{x}} \mathrm{d}x$

Since,

$\ln { ( } 1+x)=\displaystyle \sum _{ n=1 }^{ \infty } \dfrac { (-1)^{ n+1 } }{ n } x^{ n }=x-\dfrac { x^{ 2 } }{ 2 } +\dfrac { x^{ 3 } }{ 3 } -\cdots \\ \quad { for }\quad \left| x \right| \leq 1,\quad { unless }\quad x=-1.$

Which Can Be Simplified to,

$\ln { \beta }$ = $\displaystyle\int_{0 }^{1}{ \dfrac{x-\dfrac { x^{ 2 } }{ 2 } +\dfrac { x^{ 3 } }{ 3 } -\cdots}{x}} \mathrm{d}x$

Which Again Simplifies to,

$\ln { \beta }$ = $\displaystyle \int_{0 }^{1}1-\dfrac { x }{ 2 } +\dfrac { x^{2} }{ 3 } -\cdots\mathrm{d}x$

Integrating We Get,

$\ln { \beta }$ = $\displaystyle \left. x - \dfrac { x^{ 2 } }{ 2^{2} } +\dfrac { x^{ 3 } }{ 3^{2} } - \dfrac { x^{ 4} }{ 4^{2}} +\cdots \right|_0^1$

Putting The Limits We Get,

$\ln { \beta }$ = $\displaystyle 1 - \dfrac { 1 }{ 2^{2} } +\dfrac {1 }{ 3^{2} } - \dfrac {1}{ 4^{2}} +\cdots$

And It is Known That

$\displaystyle \sum _{ n=1 }^{ \infty } \dfrac { (-1)^{ n+1 } }{ n^{2} }$ = $\dfrac{\pi^{2}}{12}$

So

$\ln { \beta }$ = $\dfrac{\pi^{2}}{12}$

And

$\beta$ = ${ \mathrm{e} }^{ \dfrac{\pi ^{ 2}}{12} }$

Which Is Approximately

$\beta$ = $\textbf{2.27610....}$

Take logarithm on both sides. Sum the series using integration and evaluate the integral by using zeta function. Take antilog and that's the answer

it was good that i remembered some expansions and special series and suggest you remember them too ! :) taylor and basel problem types