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I need to find the limit of

$ \lim_{n\to\infty} \frac{1}{n} \cdot \sqrt[n]{(n+1)(n+2)...(n+n)}$

using integralsums. But I can't even get a sum. The best I could do is use a product

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

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

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

$\lim_{n\to\infty} \prod_{k = 1}^{n} \left(\frac{n\cdot(1+\frac{k}{n})}{n}\right)^\frac{1}{n} = $

$\lim_{n\to\infty} \prod_{k = 1}^{n} (1+\frac{k}{n})^\frac{1}{n} $

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  • $\begingroup$ Great start! Try taking the logarithm of your final expression. $\endgroup$
    – B. Mehta
    Nov 28, 2017 at 22:35
  • 1
    $\begingroup$ $$\exp\int_{0}^{1}\log(1+x)\,dx = \frac{4}{e}.$$ $\endgroup$ Nov 29, 2017 at 13:28

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