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What's the value of $$\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}?$$ I've tried the AM-GM inequality with no luck. Also tried to right the inequality as: $$\left(\left(2-\frac{1}{2}\right)\left(2-\frac{1}{3}\right)\dots\left(2-\frac{1}{n+1}\right)\right)^\frac{1}{n}.$$ Would be happy to hear any hints/proofs.

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  • $\begingroup$ try to take logs $\endgroup$ – tired Jun 28 '16 at 9:36
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Clearly the expression is $a_{n}^{1/n}$ where $a_{n} = \dfrac{(2n + 1)!}{n!(n + 1)!2^{n}}$ and hence $$\frac{a_{n + 1}}{a_{n}} = \frac{(2n + 3)!}{2^{n + 1}(n + 1)!(n + 2)!}\cdot\frac{2^{n}n!(n + 1)!}{(2n + 1)!} = \frac{(2n + 3)}{(n + 2)} \to 2$$ so desired limit is also $2$. We have used the following standard result:

If $\{a_{n}\}$ is a sequence of positive terms such that $a_{n + 1}/a_{n} \to L$ then $a_{n}^{1/n} \to L$.

One can also use Stirling's approximation for $n!$ but proof of Stirling's approximation is not easy.

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Hint $$\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\frac{(2n+1)!}{{{2}^{n}}n!(n+1)!}$$ Note $$\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\left( \frac{2\times 4\times 6\times \cdots \times 2n}{{{2}^{n}}(1\times 2\times 3\times \cdots \times n)} \right)\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\frac{(2n+1)!}{{{2}^{n}}n!(n+1)!}$$

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