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Find $$ \lim_{n\to \infty}\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}.$$

I tried with ln composition but ineffectively. Any idea?

Robert Z
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Marine Galantin
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  • Try to use integrals. – Michael Rozenberg Jan 08 '18 at 14:35
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    Use Cesaro-Stolz and try to find limit of $b_{n+1}/b_n$ where $b_n=a_n^n$ and $a_n$ is given sequence in question. – Paramanand Singh Jan 08 '18 at 14:35
  • @MichaelRozenberg I know that it is possible but didn't see how at school yet. But if you can explain it I d learn with pleasure :) – Marine Galantin Jan 08 '18 at 14:38
  • See https://math.stackexchange.com/questions/1598508/i-need-help-to-advance-in-the-resolution-of-that-limit-lim-n-to-infty-s/1598524#1598524 and https://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn/475791#475791 – lab bhattacharjee Jan 08 '18 at 14:39
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    $\ln\left( \prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}\right)= \frac{1}{n}\sum_{k=1}^{2n}\ln\left(1+ \frac{ k}{n} \right)$ which is a Riemann sum of $\int_0^2\ln(1+x)dx$. – Robert Z Jan 08 '18 at 14:40
  • i find one half of the integral between 1 and 1,5 of ln(t). Is it this ? – Marine Galantin Jan 08 '18 at 14:49

3 Answers3

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Following the suggestion by Paramanand Singh let indicate

$$a_n=\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}=\sqrt[n] b_n \quad b_n=\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)$$ then

$$\frac{b_{n+1}}{b_n}=\frac{\prod_{k=1}^{2n+2} \left(1+ \frac{ k}{n+1} \right)}{\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)}=\left(1+\frac{2n+2}{n+1}\right)\left(1+\frac{2n+1}{n+1}\right)\prod_{k=1}^{2n} \frac{n(n+k+1)}{(n+1)(n+k)} $$

$$=\left(1+\frac{2n+2}{n+1}\right)\left(1+\frac{2n+1}{n+1}\right)\frac{n^{2n}(3n+1)}{(n+1)^{2n}(n+1)} \to 3\cdot 3\cdot \frac{3}{e^2}=\frac{27}{e^2}$$

and therefore (see Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$)

$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} \sqrt[n] b_n =\lim_{n\to \infty} \frac{b_{n+1}}{b_n}=\frac{27}{e^2}$$

user
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    interesting method thank you ! – Marine Galantin Jan 08 '18 at 19:45
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    @MarineGalantin You are welcome, I've only write down a solution following the method indicated by Paramanand Singh. Indeed it is very useful when you deal with sequences like $\sqrt[n] {a_n}$. – user Jan 08 '18 at 20:39
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Hint:

Compute the limit of the logarithm, noticing that

$$\log\biggl(\prod_{k+1}^{2n}\Bigl(1+\frac kn\Bigr)^{\tfrac1n}\biggr)=\sum_{k+1}^{2n}\frac1n\log\Bigl(1+\frac kn\Bigr).$$ Looks very much like an upper Riemann sum.

Bernard
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We have that $$\begin{align} \prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}&= \frac{1}{n^2} \left(\prod_{k=1}^{2n}(n+ k)\right)^{1/n}=\frac{1}{n^2}\left(\frac{(3n)!}{n!}\right)^{1/n}\\&\sim \frac{1}{n^2}\left(\frac{(3n)^{3n}\sqrt{2\pi (3n)}}{e^{3n}}\cdot \frac{e^{n}}{n^{n}\sqrt{2\pi n}}\right)^{1/n}=\frac{3^3}{e^2}\left(\sqrt{3}\right)^{1/n}\to\frac{27}{e^2}\end{align} $$ where we used the Stirling approximation for the factorials.

Robert Z
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