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$$\lim_{n \rightarrow \infty}{\prod_{j=1}^{n}\Big(1+\frac{2j-1}{2n}\Big)^{\frac{1}{2n}}}$$

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

user466946
  • 49

4 Answers4

2

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

$$\frac{b_{n+1}}{b_n}=\left(\frac{(4n+3)\cdot ...\cdot (2n+5)\cdot (2n+3)}{2^{n+1}(n+1)^{n+1}} \cdot\frac{2^nn^n}{(4n-1)\cdot ...\cdot (2n+3)\cdot (2n+1)}\right)^{\frac12}=\left(\frac{(4n+3)(4n+1)}{2(n+1)(2n+1)}\cdot \frac{1}{\left(1+\frac1n\right)^n}\right)^\frac12\to \left(\frac{16}{4}\cdot \frac1e\right)^\frac12=\frac2{\sqrt e}$$

user
  • 154,566
2

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 $$ \lim_{n\to\infty}u_n=e^{-\frac12+\ln2}=\frac{2}{\sqrt e}.$$

xpaul
  • 44,000
1

As you noted $$ \begin{align} \log\left[\prod_{j=1}^n\left(1+\frac{2j-1}{2n}\right)^{\frac1{2n}}\right] &=\sum_{j=1}^n\frac1{2n}\log\left(1+\frac{2j-1}{2n}\right) \end{align} $$ 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 $$ e^{\log(2)-\frac12}=2e^{-1/2} $$

robjohn
  • 345,667
0

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 $P(n)\; \to \; 2/\sqrt{e}$

G Cab
  • 35,272
  • I obtain $\frac4{\sqrt e}$, can you check please. Thanks – user Feb 21 '18 at 21:19
  • @gimusi: direct computation shows that limit approaches $2/\sqrt{e}$, but if I understand correctly, that's also what you reach in your answer : $\sqrt{(16/4 \cdot 1/e)}$ – G Cab Feb 21 '18 at 22:22
  • oh yes of course! you have calculated the square! I fix the typo, thanks – user Feb 21 '18 at 22:23