6
$\begingroup$

Using the AM-GM inequality we get

$$\dfrac{1}{n}\sum_{i=1}^{n}\bigg(1+\dfrac{(2i-1)}{n}\bigg)>\prod_{i=1}^{n}\bigg(1+\dfrac{(2i-1)}{n}\bigg)^{{1}/{n}}\\ \implies \dfrac{n+\frac{1}{2n}(1+3+5+...(2n-1))}{n}>\prod_{i=1}^{n}\bigg(1+\dfrac{(2i-1)}{n}\bigg)^{{1}/{n}}\\ \implies \frac{n+\frac{1}{2n}\left[\frac{n}{2}\left[2\cdot 1+(n-1)2\right]\right]}{n}>\prod_{i=1}^{n}\bigg(1+\frac{(2i-1)}{n}\bigg)^{{1}/{n}}\\ \implies \dfrac{3}{2}>\prod_{i=1}^{n}\bigg(1+\dfrac{(2i-1)}{n}\bigg)^{{1}/{n}}\therefore \sqrt{\dfrac{3}{2}}>\prod_{i=1}^{n}\bigg(1+\dfrac{(2i-1)}{n}\bigg)^{{1}/{2n}}$$(By taking square root of both sides)

Then what can we say about the $$\lim_{n\to \infty}\left\{\left(1+\dfrac{1}{2n}\right)\cdot \left(1+\dfrac{3}{2n}\right) \cdot \left(1+\dfrac{5}{2n}\right)\cdots\left(1+\dfrac{2n-1}{2n}\right)\right\}^{{1}/{2n}}$$

@Jair Taylor has mentioned the usage of Reimann sums, I am going to try it here.

$$\lim_{n\to\infty}\sum_{i=1}^{n}\log\left(1+\dfrac{2i-1}{n}\right)\cdot \dfrac{1}{n}$$

$$\therefore a=1, \Delta x_i=\dfrac{1}{n} \therefore \dfrac{1}{n}=\dfrac{b-1}{n} \therefore b=2$$

Hence the integral becomes $\int_{1}^2(\log(?)dx$

What do I put here instead of '?' or is the approach wrong?

But again just see another thing $\int_{1}^2(x)dx=\dfrac{3}{2}$ and we already have $\sqrt{\dfrac{3}{2}}$ from our inequality!!!

$\endgroup$
2
  • 3
    $\begingroup$ If you take a log it looks a bit like a Riemann sum. $\endgroup$ Apr 27, 2018 at 17:35
  • $\begingroup$ Yes @JairTaylor thats a good way of approach. $\endgroup$
    – Saradamani
    Apr 27, 2018 at 17:38

3 Answers 3

5
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\lim_{n \to \infty}\bracks{\prod_{k = 1}^{n} \pars{1 + {2k - 1 \over 2n}}}^{1/\pars{2n}}}} = \lim_{n \to \infty}\bracks{{1 \over n^{n}}\prod_{k = 1}^{n} \pars{k + n - {1 \over 2}}}^{1/\pars{2n}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{{1 \over n^{n}} \pars{{1 \over 2} + n}^{\large\overline{n}}}^{1/\pars{2n}} = \lim_{n \to \infty}\bracks{{1 \over n^{n}} {\Gamma\pars{1/2 + 2n} \over \Gamma\pars{1/2 + n}}}^{1/\pars{2n}} \\[5mm] = &\ \lim_{n \to \infty}\bracks{{1 \over n^{n}} {\root{2\pi}\pars{2n - 1/2}^{2n}\expo{-2n + 1/2} \over \root{2\pi}\pars{n - 1/2}^{n}\expo{-n + 1/2}}}^{1/\pars{2n}} \\[5mm] = &\ \lim_{n \to \infty}\braces{{1 \over n^{n}} {\pars{2n}^{2n}\bracks{1 - \pars{1/2}/\pars{2n}}^{\, 2n} \over n^{n}\bracks{1 - \pars{1/2}/n}^{n}}\,\expo{-n}}^{1/\pars{2n}} \\[5mm] = &\ \lim_{n \to \infty}\braces{2^{2n}\, {\bracks{1 - \pars{1/2}/\pars{2n}}^{\, 2n} \over \bracks{1 - \pars{1/2}/n}^{n}}\,\expo{-n}}^{1/\pars{2n}} = \lim_{n \to \infty}\pars{2^{2n}\expo{-n}}^{1/\pars{2n}} \\[5mm] = &\ \bbx{2 \over \root{\expo{}}} \approx 1.2131 \end{align}

$\endgroup$
4
$\begingroup$

I'll give it a try with squeezing and the follwing Stirling approximation:

  • $\sqrt{2\pi}\sqrt{n}\frac{n^n}{e^n} \leq n! \leq e \sqrt{n}\frac{n^n}{e^n}$

Let

  • $p_n = \prod_{k=1}^n\left(1 + \frac{2k-1}{2n} \right)$
  • $P_n = (p_n)^{\frac{1}{2n}}$

We get $$\prod_{k=1}^n\left(1 + \frac{k-1}{n} \right) =\prod_{k=1}^n\left(1 + \frac{2k-2}{2n} \right) \leq p_n \leq \prod_{k=1}^n\left(1 + \frac{2k}{2n} \right) = \prod_{k=1}^n\left(1 + \frac{k}{n} \right)$$ Hence, $$\frac{n}{2n}\frac{(2n)!}{n!\cdot n^n}=\frac{\prod_{k=1}^n(n+k-1)}{n^n} \leq p_n \leq \frac{\prod_{k=1}^n(n+k)}{n^n}=\frac{(2n)!}{n!\cdot n^n}$$ Now, we use the Stirling approximations on both sides:

$$\frac{1}{2}\frac{\sqrt{2\pi}\cdot \sqrt{2n}\cdot (2n)^{2n}\cdot e^n}{e\cdot \sqrt{n}\cdot n^n\cdot e^{2n} \cdot n^n} \leq p_n \leq \frac{e\cdot \sqrt{2n}\cdot (2n)^{2n}\cdot e^n}{\sqrt{2\pi}\cdot \sqrt{n}\cdot n^n\cdot e^{2n} \cdot n^n}$$ Collecting the constant factors in positive constants $c, C$: $$c\frac{2^{2n}}{e^n}\leq p_n \leq C\frac{2^{2n}}{e^n}$$ So, we get $$c^{\frac{1}{2n}}\frac{2}{\sqrt{e}} \leq P_n \leq C^{\frac{1}{2n}}\frac{2}{\sqrt{e}}$$ $$\lim_{n\rightarrow \infty}P_n = \frac{2}{\sqrt{e}}$$

$\endgroup$
2
  • $\begingroup$ Is there any easier way? @trancelocation as this method is understandable but I have had heard of this Sterling's Approximation only once in my life long ago. $\endgroup$
    – Saradamani
    Apr 27, 2018 at 19:24
  • 1
    $\begingroup$ @Saradamani: Hmmm. That's a good question. I tried it first via logarithms but it didn't work. So, I tried to squeeze the product between nicer products. And when factorials pop up as here then Stirling and Gamma are not far away. Maybe someone else can produce a more elementary way of calculating the limit. $\endgroup$ Apr 27, 2018 at 19:34
2
$\begingroup$

hint: $P_n = \dfrac{\sqrt[2n]{(1+1/2n)(1+2/2n)(1+3/2n)\cdots (1+2n/2n)}}{\sqrt[2n]{(1+2/2n)(1+4/2n)(1+6/2n)\cdots (1+2n/2n)}}$, and take log and you have $\log(P_n) = R_n - Q_n$, whereas $R_n, Q_n$ are the Riemann sums.

$\endgroup$
1
  • $\begingroup$ I am not quite conversant with Reimann Sums. Can you please show me how. Thanks! $\endgroup$
    – Saradamani
    Apr 27, 2018 at 17:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .