Find the limit of the given function 
$$\lim_{n \rightarrow \infty}{\prod_{j=1}^{n}\Big(1+\frac{2j-1}{2n}\Big)^{\frac{1}{2n}}}$$

I actually used logarithm after equating it with $l$.
That is
$$\log{l}=\frac{1}{2n}\sum_{j=1}^{n}\log{\Big(1+\frac{2j-1}{2n}\Big)}$$
After that, I was unable to proceed
 A: Let refer to ratio-root criteria with


*

*$a_n=\prod_{k=1}^{n} \left(1+\frac{2k-1}{2n}\right)^{\frac1{2n}}=(b_n)^\frac1n$

*$b_n=\prod_{k=1}^{n} \left(1+\frac{2k-1}{2n}\right)^{\frac12}= \left(\frac{(4n-1)\cdot ...\cdot (2n+3)\cdot (2n+1)}{2^nn^n}\right)^{\frac12}$


then
$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$
and since
$$\frac{b_{n+1}}{b_n}=\left(\frac{(4n+3)\cdot ...\cdot (2n+5)\cdot (2n+3)}{2^{n+1}(n+1)^{n+1}}
\cdot\frac{2^nn^n}{(4n-1)\cdot ...\cdot (2n+3)\cdot (2n+1)}\right)^{\frac12}=\left(\frac{(4n+3)(4n+1)}{2(n+1)(2n+1)}\cdot \frac{1}{\left(1+\frac1n\right)^n}\right)^\frac12\to \left(\frac{16}{4}\cdot \frac1e\right)^\frac12=\frac2{\sqrt e}$$
A: You can use the Riemann sum to get the limit. Let
$$ u_n=\prod_{j=1}^{n}\Big(1+\frac{2j-1}{2n}\Big)^{\frac{1}{2n}} $$
and then
$$ \ln u_n=\frac1{2n}\sum_{j=1}^{n}\ln\Big(1+\frac{2j-1}{2n}\Big). $$
Thus
\begin{eqnarray}
\lim_{n\to\infty}\ln u_n&=&\lim_{n\to\infty}\frac1{2n}\sum_{j=1}^{n}\ln\Big(1+\frac{2j-1}{2n}\Big)\\
&=&\lim_{n\to\infty}\left[\frac1{2n}\sum_{j=1}^{n}\ln\Big(1+\frac{2j-1}{2n}\Big)+\frac1{2n}\sum_{j=1}^{n}\ln\Big(1+\frac{2j}{2n}\Big)\right]-\lim_{n\to\infty}\frac1{2n}\sum_{j=1}^{n}\ln\Big(1+\frac{2j}{2n}\Big)\\
&=&\lim_{n\to\infty}\frac1{2n}\sum_{j=1}^{2n}\ln\Big(1+\frac{j}{2n}\Big)-\lim_{n\to\infty}\frac1{2n}\sum_{j=1}^{n}\ln\Big(1+\frac{j}{n}\Big)\\
&=&\int_0^1\ln(1+x)dx-\frac12\int_0^1\ln(1+x)dx\\
&=&\frac12\int_0^1\ln(1+x)dx\\
&=&-\frac12+\ln2
\end{eqnarray}
and so
$$ \lim_{n\to\infty}u_n=e^{-\frac12+\ln2}=\frac{2}{\sqrt e}.$$
A: As you noted
$$
\begin{align}
\log\left[\prod_{j=1}^n\left(1+\frac{2j-1}{2n}\right)^{\frac1{2n}}\right]
&=\sum_{k=1}^n\frac1{2n}\log\left(1+\frac{2j-1}{2n}\right)
\end{align}
$$
The sum on the right is a Riemann sum for
$$
\frac12\int_0^1\log(1+x)\,\mathrm{d}x=\log(2)-\frac12
$$
Therefore, the limit of the product is
$$
e^{\log(2)-\frac12}=2e^{-1/2}
$$
A: In another way
$$
\eqalign{
  & P(n) = \prod\limits_{j = 1}^n {\left( {1 + {{2j - 1} \over {2n}}} \right)^{\,1/\left( {2n} \right)} }  = \prod\limits_{j = 1}^n {\left( {{{n - 1/2 + j} \over n}} \right)^{\,1/\left( {2n} \right)} }  =   \cr 
  &  = \prod\limits_{j = 0}^{n - 1} {\left( {{{n + 1/2 + j} \over n}} \right)^{\,1/\left( {2n} \right)} }  \cr} 
$$
squaring, converting to Rising Factorial expressed through the Gamma function
$$
P^{\,2} (n) = {1 \over n}\left( {\left( {n + 1/2} \right)^{\,\overline {\,n\,} } } \right)^{\;1/n}  = {1 \over n}\left( {{{\Gamma \left( {2n + 1/2} \right)} \over {\Gamma \left( {n + 1/2} \right)}}} \right)^{\;1/n} 
$$
and then using Stirling approximation
$$
\eqalign{
  & P^{\,2} (n) \approx {1 \over n}\left( {{{\sqrt {\,{{2\,\pi } \over {2n + 1/2}}\,} \left( {{{2n + 1/2} \over e}} \right)^{2n + 1/2} } \over {\sqrt {\,{{2\,\pi } \over {n + 1/2}}\,} \left( {{{n + 1/2} \over e}} \right)^{n + 1/2} }}} \right)^{\;1/n}  \approx   \cr 
  &  \approx {1 \over n}\left( {{{\left( {2n + 1/2} \right)^{2n + 1/2} } \over {\sqrt {\,2\,} e^n \left( {n + 1/2} \right)^{n + 1/2} }}} \right)^{\;1/n}  \approx {n \over n}\left( {\left( 2 \right)^{2n + 1/2} {{\left( {1 + 1/2n} \right)^{n + 1/2} } \over {\sqrt {\,2\,} e^n }}} \right)^{\;1/n}  \approx   \cr 
  &  \approx {4 \over e} \cr} 
$$
That is $P(n)\; \to \; 2/\sqrt{e}$
