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

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

It is a nice proof, and I do not think it can be shortened, unless some very ingenious manipulation is eluding me. — Jack D'Aurizio, Sep 18 '15 at 12:20
Note that $\int_0^1 x\ln x,dx=-{1\over 4}$ so the final result is $\sqrt{e}$. — uniquesolution, Sep 18 '15 at 12:27
Probably you can leave out everything before $$\ln y = \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n}_{r=1}\frac{r}{n}\ln\left(1+\frac{n}{r}\right)$$. — GEdgar, Sep 18 '15 at 13:05
Yeah, the final answer is wrong, so I think you have an error in your integrals. — Thomas Andrews, Sep 18 '15 at 13:16
Thanks Jack D'Aurizio, Thanks uniquesolution I have edited it. — juantheron, Sep 18 '15 at 13:38

Saul Mendoza · Accepted Answer · 2015-12-09T08:53:44.757

2

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: $$ \ln y = \lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^{n}_{r=1} r\ln\left(1+\frac{n}{r}\right)\\~\\={\lim_{n\rightarrow \infty}\frac{1}{2n+1}\left[\sum^{n+1}_{r=1} r\ln\left(1+\frac{n+1}{r}\right)- \sum^{n}_{r=1} r\ln\left(1+\frac{n}{r}\right)\right]}\\~\\ \geq \lim_{n\rightarrow \infty}\frac{1}{2n+1}\left[ (n+1)\ln\left(1+\frac{n}{n+1}\right)\right]\\~\\={\lim_{n\rightarrow \infty}\frac{n}{2n+1}\ln\left(1+\frac{n}{n+1}\right)^{\frac{n+1}{n}}}\\~\\=\frac{1}{2}*ln(e)=\frac {1}{2} $$

edited Dec 09 '15 at 08:53

answered Dec 09 '15 at 08:48

Saul Mendoza

231

I hope to be helpful, i understood you were looking for an alternative. – Saul Mendoza Dec 09 '15 at 08:49
In a similar fashion, it can be demonstrated that the upper bound is 1/2, so that your limit is pinched to 1/2 – Saul Mendoza Dec 09 '15 at 08:52

Evaluation of $ \lim_{n\rightarrow \infty}\left[\prod^{n}_{r=1}\left(1+\frac{n}{r}\right)^{\frac{r}{n}}\right]^{\frac{1}{n}}$

1 Answers1