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Find the limit of the following sequence.

$$a_n = \frac{\sqrt[n]{(n+1)(n+2)...(2n)}}{n}$$

I tried couple of methods: Stolz, Squeeze, D'Alambert. But I can not seem to make a conclusion on the limit.

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3 Answers 3

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Consider the logarithm of $a_n$ $$\log a_n = \frac{1}{n} \sum_{k=1}^n \log \left( 1+ \frac{k}{n}\right)$$ You can recognize that this is a Riemann sum of the integral $\int_0^1 \log (1+x) dx = 2 \log 2 -1$.

Hence the limit evaluates to $$e^{2 \log 2 -1} = \frac 4e$$

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  • $\begingroup$ I am the one who wrote the answer that "does not make sense". I'm curious about the reason. Why those approximations are wrong? Thankyou $\endgroup$
    – Who knows
    Jan 27, 2016 at 19:43
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$n! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^{n} \implies (2n)! \sim \sqrt{4\pi n} \left( \frac{2n}{e} \right)^{2n}$

\begin{align*} (n+1)(n+2) \ldots (2n) &= \frac{(2n)!}{n!} \\ & \sim \sqrt{2} \left( \frac{4n}{e} \right)^{n} \\ \lim_{n\to \infty} \frac{\displaystyle{\sqrt[n]{(n+1)(n+2) \ldots (2n)}}}{n} &= \lim_{n\to \infty} \left( \frac{\sqrt[2n]{2}}{n} \times \frac{4n}{e} \right) \\ &= \frac{4}{e} \end{align*}

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We can use the following lemma:

Lemma: let $a_n$ be a sequence with positive terms. If $\lim_{n\to >\infty}=\frac{a_{n+1}}{a_n}=l$, then $\lim_{n\to \infty}\sqrt[n]{a_n}=l$.

Proof: suppose $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=l$. Then for every $\varepsilon>0$ there exists $N_{\varepsilon}$ such that for every $n\geq N_{\varepsilon}$ $$\left |\frac{a_{n+1}}{a_n}-l \right |<\varepsilon$$ So $$|a_n|=\frac{|a_n|}{|a_{n-1}|}\frac{|a_{n-1}|}{|a_{n-2}|}\cdots\frac{|a_{N+1}|}{|a_N|}<(l+\varepsilon)^{n-N}|a_N|$$ Considering the $n$-th root of both sides, we have $$\sqrt[n]{|a_n|}<(l-\varepsilon)^{\frac{n-N}{n}}\sqrt[n]{|a_N|}$$ Taking its limit for $n\to \infty$, we obtain $$\sqrt[n]{|a_n|}\le l+\varepsilon $$ One can show analogously that $\sqrt[n]{|a_n|}\ge l-\varepsilon$

Consider the sequence $$a_n=\frac{(n+1) \cdot \ldots \cdot (2n)}{n^n}$$

The ratio between two consequent terms is $$\frac{a_{n+1}}{a_n}=\frac{(n+2)\cdot \ldots \cdot (2n)(2n+1)(2n+2)}{(n+1)(n+1)(n+2)\cdot \ldots \cdot (2n)}\cdot \left (\frac{n}{n+1} \right )^n=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}\left (1+\frac 1n \right )^{-n}$$ So $$\lim_{n \to \infty}\frac{a_{n+1}}{{a_n}}=\frac{4}{e}$$ By the previous lemma,

$$\lim_{n\to \infty}\frac{\sqrt[n]{(n+1) \cdot \ldots \cdot (2n)}}{n}=\frac{4}{e}$$

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