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Im a new participant in this mathematical forum, so this is one of that i couldn't solve it.

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

I've tried to transform the product in a summation as function of a logarithmic function, and I wasn't been successful. Like $u=x^n$, $du=nx^{n-1}dx$

$g_n(u) = \sqrt[n]{\frac{ dx}{du} \cdot x^{\frac{n(n+1)}{2}} \cdot \prod_{k=1}^n(1 - x^k)}$

$g_n(u)=\sqrt[n]{\frac{1}{nx^{n-1}}} \cdot x^{\frac{n+1}{2}} \cdot e^{\frac 1n\sum_{k=1}^n \ln(1 - x^k)}$

dammat
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2 Answers2

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See if this converts to a Riemann sum which converges to an integral: $$g_n(u)=\sqrt[n]{\frac{1}{nx^{n-1}}} \cdot x^{\frac{n+1}{2}} \cdot e^{\frac 1n\sum_{k=1}^n \ln(1 - x^k)}=\sqrt[n]{\frac{1}{nx^{1-1/n}}} \cdot u^{\frac{n+1}{2n}} \cdot e^{\frac 1n\sum_{k=1}^n \ln(1 - x^k)}\\\sim \sqrt{u}\exp\int_0^1\ln(1-u^t)dt=\sqrt u\exp\frac{\zeta(2)-{\rm Li}_2(u)}{\ln u}$$ The maxima seems to be $\approx0.185155$ at $u\approx 0.245254$ without possibly a closed form as wolfram says.

RE60K
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    (+1) I agree. The limit is the value of $$\sqrt{u}\exp\frac{\zeta(2)-\operatorname{Li}_2(u)}{\log u}$$ in the first positive root of $$\zeta(2)-\operatorname{Li}_2(u) = \frac{1}{2}\log^2 u+\log u \log(1-u), $$ hence $u(1-u)$ for $u\approx 0.245254$. – Jack D'Aurizio Jan 29 '15 at 15:44
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As $$\sqrt{x^k(1-x^k)}\le\frac{x^k+(1-x^k)}2=\frac12\iff x^k(1-x^k)\le\frac14$$ So: $$\lim_{n \to \infty } \sqrt[n]{\int_0 ^1 x^{\frac{n(n+1)}{2}}(1-x)(1-x^2)\cdots(1-x^n)d x}=\lim_{n \to \infty } \sqrt[n]{\int_0 ^1 \prod_{k=1}^nx^k(1-x^k)}\le\lim_{n\to\infty}\sqrt[n]{\int_0^1\left(\frac14\right)^ndx}=\frac14$$ So the limit is $\displaystyle\le\frac14$ Note that equality holds here not for a definite values but $\sqrt[k]2$ which are all less than $0.5$

RE60K
  • 18,045