Calculation of $\lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}$

Question

Calculation of $\displaystyle \lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}$

$\bf{My\; Solution::}$

Let $y=\displaystyle \lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}$, Now taking $\log_{e}$ on both side

$\displaystyle \log_{e}(y) = \lim_{n\to \infty}\log_{e}\left(\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}\right)=\lim_{n\to \infty}\frac{1}{n}\cdot \sum_{r=1}^{n}\log\left(1+\frac{r}{n}\right)$

Now Using Reinman Sum of Integral ...

$\displaystyle \log_{e}(y) = \lim_{n\to\infty}\frac{1}{n}\cdot \sum_{r=1}^{n}\log\left(1+\frac{r}{n}\right) = \int_{0}^{1}\ln(1+x)dx = \log_{e}(4)-\log_{e}(e) = \log_{e}\left(\frac{4}{e}\right)$

So $\displaystyle y =\lim_{n\to \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}} =\frac{4}{e}$

But how can i solve above question using Stirling Approximation,

May I ask for help?

Thank you in advance!

Answer 1

$$\left(\frac{(2n)!}{n!n^n}\right)^{1/n}\approx\left[\left(\frac{2n}{e}\right)^{2n}\left(\frac{e}{n}\right)^n\frac{1}{n^n}\right]^{1/n}$$ together with powers of $\pi$ that don't survive the $n$th root.

Answer 2

Are you allowed to use Stirling's? Rewrite everything as $$ S_n = \bigg(\binom{2n}{n} \frac{(n!)^2}{n^n} \bigg)^{\frac{1}{n}} $$ and use it , and then take the limit