Find $\lim_{n\to \infty}\{(1+\frac{1}{2n})\cdot (1+\frac{3}{2n}) \cdot (1+\frac{5}{2n})\cdots(1+\frac{2n-1}{2n})\}^{{1}/{2n}}$ 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!!!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\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}
A: 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 
$$c^{\frac{1}{2n}}\frac{2}{\sqrt{e}} \leq P_n \leq C^{\frac{1}{2n}}\frac{2}{\sqrt{e}}$$
$$\lim_{n\rightarrow \infty}P_n = \frac{2}{\sqrt{e}}$$
A: 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.
