Evaluate $\lim_{n\rightarrow\infty}$ $\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$

Question

Question: Evaluate lim$_{n\rightarrow\infty}$$\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$

My Approach Let $a_{n}=\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}$

$$a_{n+1}=\frac{\left(n+2\right)\left(n+3\right).....\left(2n+2\right)}{\left(n+1\right)^{\left(n+1\right)}}$$

$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=\frac{4}{\exp (1)}$$

Using Cauchy's second theorem On Limits

Lim$_{n\rightarrow\infty}\left[\frac{a_{2}}{a_{1}}\frac{a_{3}}{a_{2}}..........\frac{a_{n+1}}{a_{n}}\right]^{\frac{1}{n}}$=Lim$_{n\rightarrow\infty}$$\left[\frac{a_{n+1}}{a_{1}}\right]^{\frac{1}{n}}=\frac{4}{\exp(1)}$

Now I don't know how to move forward

You are almost done. Just fix the use of Cauchy theorem and say that using Cauchy's theorem we have $\lim a_n^{1/n}=\lim (a_{n+1}/a_n)=4/e$ — Paramanand Singh, Jan 01 '18 at 11:36
I think he made a typo and went ahead with it. The integral needed is $\int_{0}^{1}\log(1+x),dx$ — Paramanand Singh, Jan 01 '18 at 11:39
See also: https://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn and https://math.stackexchange.com/questions/1598508/i-need-help-to-advance-in-the-resolution-of-that-limit-lim-n-to-infty-s — lab bhattacharjee, Jan 01 '18 at 11:57
@Rick Almost! Still now .i just don't how to prove lim $\left(\frac{a_{n+1}}{a_{1}}\right)^{\frac{1}{n}}$=lim$\left(a_{n}\right)^{\frac{1}{n}}$ — Mohan Sharma, Jan 01 '18 at 12:04

jonsno · Answer 1 · 2018-01-01T11:42:10.183

3

Let the limit be $L$ then:

$$\begin{align} \ln(L) &= \lim_{n\to \infty}\frac{1}{n} \sum_{i=1}^{n} \ln\left( 1 + \frac{i}{n} \right) \\ &= \int_{0}^{1}\ln( 1+x ) dx \\ &= (x+1)\ln(1+x) -x |_{0}^{1} \\ &= 2\ln(2) - 1 \end{align}$$

So the required limit is $\exp({2\ln (2) -1}) = \color{blue}{\dfrac{4}{e}}$

edited Jan 01 '18 at 11:42

answered Jan 01 '18 at 11:31

jonsno

7,521

It's a typo I guess. You forgot the logs. You need integral of $\log(1+x)$. – Paramanand Singh Jan 01 '18 at 11:38
@paramanandSingh Ah yes forgot to put the log! Thanks – jonsno Jan 01 '18 at 11:40
Ok my +1 delivered to you. – Paramanand Singh Jan 01 '18 at 11:40
@ParamanandSingh I have a similar question which I post here: question. Thank you for thoughts :) – jonsno Jan 01 '18 at 11:55

score 3 · Accepted Answer · answered Jan 01 '18 at 11:40

$$\begin{eqnarray*}\lim_{n\to +\infty}\left[\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right]^{1/n}&=&\exp\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\\&=&\exp\int_{0}^{1}\log(1+x)\,dx=\color{red}{\frac{4}{e}}.\end{eqnarray*} $$

Evaluate $\lim_{n\rightarrow\infty}$ $\left[\frac{\left(n+1\right)\left(n+2\right).....\left(n+n\right)}{n^{n}}\right]^{\frac{1}{n}}$

2 Answers2