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$$ \lim_{n\to\infty}\sqrt[n]{\int_0^1x^{\frac{n(n+1)}{2}}(1-x)(1-x^2)\cdots(1-x^n)dx} $$

$$ =\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{4i(i+1)} $$

$$ =\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{4}(\frac{1}{i}-\frac{1}{i+1}) $$

$$ =\lim_{n\to\infty}\frac{1}{4}(1-\frac{1}{n+1}) $$

$$ =\lim_{n\to\infty}\frac{1}{4}(\frac{n}{n+1}) $$

$$ =\frac{1}{4} $$

I found an answer, but I never understood where this first step came from.

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1 Answer 1

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Using $AM\ge GM$

$$\frac{x^t+(1-x^t)}2\ge\sqrt{x^t(1-x^t)}$$ $$\frac14 \ge x^t(1-x^t)$$

$$\lim_{n \to \infty } \sqrt[n]{\int_0 ^1 x^{\frac{n(n+1)}{2}}(1-x)(1-x^2)\cdots(1-x^n)d x}=\lim_{n \to \infty } \sqrt[n]{\int_0 ^1 x(1-x)x^2(1-x^2)\cdots x^n(1-x^n)d x}=\lim_{n \to \infty } \sqrt[n]{\int_0 ^1 \prod_{k=1}^nx^k(1-x^k)\,dx}\le\lim_{n\to\infty}\sqrt[n]{\int_0^1\left(\frac14\right)^ndx}=\boxed{\frac14}$$

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    $\begingroup$ Nice solution. I wonder if there is any way to find more than the limit. Any idea about a possible asymptotics ? $\endgroup$ Commented Nov 6, 2023 at 14:29

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