TITLE: Evaluation of $ \lim_{n\rightarrow \infty}\left[\prod^{n}_{r=1}\left(1+\frac{n}{r}\right)^{\frac{r}{n}}\right]^{\frac{1}{n}}$
QUESTION [9 upvotes]: Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\left[(1+n)^{\frac{1}{n}}\cdot \left(1+\frac{n}{2}\right)^{\frac{2}{n}}\cdot \left(1+\frac{n}{3}\right)^{\frac{3}{n}}.........2\right]^{\frac{1}{n}}$
$\bf{My\; Try::}$ Let $$ y = \lim_{n\rightarrow \infty}\left[(1+n)^{\frac{1}{n}}\cdot \left(1+\frac{n}{2}\right)^{\frac{2}{n}}\cdot \left(1+\frac{n}{3}\right)^{\frac{3}{n}}.........2\right]^{\frac{1}{n}}$$
Now taking $$ \ln y = \lim_{n\rightarrow \infty}\frac{1}{n}\cdot \ln\left[(1+n)^{\frac{1}{n}}\cdot \left(1+\frac{n}{2}\right)^{\frac{2}{n}}\cdot \left(1+\frac{n}{3}\right)^{\frac{3}{n}}.........2\right]$$
So we get $$\ln y = \lim_{n\rightarrow \infty}\frac{1}{n}\left[\frac{1}{n}\cdot \ln(1+n)+\frac{2}{n}\ln\left(1+\frac{n}{2}\right)+........+\frac{n}{n}\ln\left(1+\frac{n}{n}\right)\right]$$
So $$\ln y = \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n}_{r=1}\frac{r}{n}\ln\left(1+\frac{n}{r}\right)$$
Now Convertinto Reinmann Sum of Integral
So put $\displaystyle \frac{r}{n} = x\;,$ Then $\displaystyle \frac{1}{n}=dx$ and Calculate limit
So we get $$\ln y = \int_{0}^{1}x\cdot \ln\left(1+\frac{1}{x}\right)dx=\int_{0}^{1}x\cdot \left[\ln(1+x)-\ln x\right]dx$$
So we get $$\ln y = \int_{0}^{1}x\cdot \ln(x+1)dx-\int_{0}^{1}x\ln xdx$$
Now after Integrate we wil get $$\ln y = \frac{1}{2}\ln 2+\frac{1}{4}-\frac{1}{2}\ln 2+\frac{1}{4} = \frac{1}{2}$$
So we get $$\ln y = \frac{1}{2}\Rightarrow y = e^{\frac{1}{2}}=\sqrt{e}$$
Can we solve it any Shorter way, If yes then plz explain here
Thanks

REPLY [2 votes]: We could use Stolz theorem, which states:
 $$\lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\ell\implies 
\lim_{n \to \infty} \frac{a_n}{b_n}=\ell$$
Now we define $a_n$ and $b_n$ as follows:
$$a_n={\sum^{n}_{r=1}r\ln\left(1+\frac{n}{r}\right)} \quad b_n=n^2$$
So that:
$$
\ln y = \lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^{n}_{r=1} r\ln\left(1+\frac{n}{r}\right)\\~\\={\lim_{n\rightarrow \infty}\frac{1}{2n+1}\left[\sum^{n+1}_{r=1} r\ln\left(1+\frac{n+1}{r}\right)-  \sum^{n}_{r=1} r\ln\left(1+\frac{n}{r}\right)\right]}\\~\\ \geq \lim_{n\rightarrow \infty}\frac{1}{2n+1}\left[ (n+1)\ln\left(1+\frac{n}{n+1}\right)\right]\\~\\={\lim_{n\rightarrow \infty}\frac{n}{2n+1}\ln\left(1+\frac{n}{n+1}\right)^{\frac{n+1}{n}}}\\~\\=\frac{1}{2}*ln(e)=\frac {1}{2}
$$