I want to find the limit of following sequence by using Cauchy theorem and I don't know how.
$\displaystyle{% \lim_{n \to \infty}\left[% \left(n + 1\right)\left(n + 2\right)\cdots\left(n + n\right) \over n^{n}\right]^{1/n}} $
Can someone help me ?.
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Felix Marin
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user378334
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2What is supposed to be the statement of the Cauchy theorem you are referring to? – Jack D'Aurizio Oct 13 '16 at 18:50
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By continuity of $\log$/$\exp$ and Riemann sums, $$\lim_{n\to +\infty}\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)^{1/n}=\exp\int_{0}^{1}\log(1+x),dx = \frac{4}{e}.$$ – Jack D'Aurizio Oct 13 '16 at 18:51
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Jack would you please solve it by using cauchy theorem on geometric mean for the sequence. – user378334 Oct 13 '16 at 18:55
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I am not aware of such theorem. By the AM-GM inequality, the limit, if existing, is $< \frac{3}{2}$. – Jack D'Aurizio Oct 13 '16 at 18:58
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The formula which you have used, is it general formula that can be applied to any such sequence? – user378334 Oct 13 '16 at 19:00
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If $f(x)$ is a continuous, positive and bounded function on the interval $(0,1)$, $$\lim_{n\to +\infty}\prod_{k=1}^{n} f(k/n)^{1/n}=\exp\int_{0}^{1}\log f(x),dx, $$ yes. – Jack D'Aurizio Oct 13 '16 at 19:05
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There are several theorems called Cauchy theorem. – Martin Sleziak Oct 13 '16 at 21:41
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1You can find the same limit here (see also linked posts) and here (see also linked posts. – Martin Sleziak Oct 13 '16 at 21:42
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By Cauchy's second limit theorem (with $a_n > 0$)
$$\lim_{n \to \infty} a_n ^{1/n} = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}. $$
In this case,
$$\begin{align} \lim_{n \to \infty}\left[\frac{(n+1)(n+2)\cdots(n+n)}{n^n}\right]^{1/n} &= \lim_{n \to \infty}\left[\frac{(2n)!}{n^n n!} \right]^{1/n} \\ &= \lim_{n \to \infty} \frac{(2n+2)!}{(n+1)^{n+1} (n+1)!} \frac{n^n n!}{(2n)!} \\ &= \lim_{n \to \infty} \frac{(2n+2)(2n+1)}{(n+1)(n+1)} \frac{n^n}{(n+1)^n} \\ &= \lim_{n \to \infty} \frac{(2+2/n)(2+1/n)}{(1+1/n)(1+1/n)} \lim_{n \to \infty} \frac{1}{(1 + 1/n)^n} \\ &= \frac{4}{e}\end{align} $$
RRL
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