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Evaluate $$\lim_{n\to\infty} \frac{1}{n} \left( (n+1)\cdots (n+n) \right)^\frac{1}{n}$$

I somehow need to get a Riemann sum from this expression in order to evaluate the limit. I tried to use $f(x)^{g(x)}= e^{\ln(f(x))g(x)}$ identity but got stuck.

$$ e^{ \ln(\frac{1}{n}(n+1)^\frac{1}{n}\cdots (n+n)^\frac{1}{n})} $$

Then, $$\lim_{n\to\infty} \ln(\frac{1}{n}(n+1)^\frac{1}{n}\cdots (n+n)^\frac{1}{n}) = \lim_{n\to\infty} \ln(\frac{1}{n}) + \sum_{i=1}^n \ln(i+1)^\frac{1}{n}$$

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  • $\begingroup$ I would start by looking at the logarithm of the expression which basically leaves you with $1/n$ times a sum of logarithms. $\endgroup$ – Dr_Be Jan 18 '16 at 9:22
  • $\begingroup$ You can take the $\frac 1n=\left(\frac 1{n^n}\right)^{\frac 1n}$ inside the product before doing anything else, if that helps. $\endgroup$ – Mark Bennet Jan 18 '16 at 9:25
  • $\begingroup$ The product $(n+1) \dots (n+n)$ is the ratio of two $\Gamma$ functions, which can be reduced to $\Gamma(2 n)/\Gamma(n)$. Then use dlmf.nist.gov/5.5.E5 and Stirling's formula. $\endgroup$ – Johannes Trost Jan 18 '16 at 9:27
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$\frac{1}{n} \left( (n+1)\cdots (n+n) \right)^\frac{1}{n}=\frac{1}{n}(n^n(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n}))^{1/n}=((1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n}))^{1/n}$.

Then $((1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{n}{n}))^{1/n}=e^{\frac{1}{n}\sum_{k=1}^n\log(1+\frac{k}{n})}\to e^{\int_0^1\log(1+x)dx}$, which is easy to calculate.

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