TITLE: how to compute this limit
QUESTION [7 upvotes]: compute $I=\lim\limits_{n\to+\infty}\sqrt[n]{\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx}$

attempt: I tried to evaluate the integral
$$\begin{align}
\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx&=\int\limits_0^1x^{n+1}\left(1-x-x^2-x^3+x^5+\cdots\right)dx\\
&=\int\limits_0^1x^{n+1}-x^{n+2}-x^{n+3}-x^{n+4}+x^{n+6}+\cdots dx\\
&=\left.\frac{x^{n+2}}{n+2}-\frac{x^{n+3}}{n+3}-\frac{x^{n+4}}{n+4}-\frac{x^{n+5}}{n+5}+\frac{x^{n+6}}{n+6}+\cdots\right|_0^1\\
&=\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}-\frac{1}{n+5}+\frac{1}{n+6}+\cdots
\end{align}$$
attempt 2:
I think I can use mean value theorem for integrals

if $f$ are continuous into $[a,b]$ then exists $\xi\in[0,1]$ such that $\int\limits_a^b f(x)dx=f(\xi)(b-a)$

then using it for $[a,b]=[0,1]$ since the function inside integral are continuous, then exists $\xi\in[0,1]$ such that 
$$\begin{align}
\sqrt[n]{\int\limits_0^1x^{n+1}(1-x)\cdots(1-x^n)dx}&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)(1-0)}\\
&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}
\end{align}$$
then making for $n\in\mathbb{N}^*$
$$\begin{align}
a_n&=\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}\\
a_{n+1}&=\sqrt[n+1]{\xi^{n+2}(1-\xi)\cdots(1-\xi^n)(1-\xi^{n+1})}\\
&=\sqrt[n+1]{\xi\xi^{n+1}(1-\xi)\cdots(1-\xi^n)(1-\xi^{n+1})}\\
&=\sqrt[n+1]{\xi\left[\sqrt[n]{\xi^{n+1}(1-\xi)\cdots(1-\xi^n)}\right]^n(1-\xi^{n+1})}\\
&=\sqrt[n+1]{a_n^n\xi(1-\xi^{n+1})}\\
a_{n+1}^{n+1}&=a_n^n\xi(1-\xi^{n+1})
\end{align}$$
and $\xi\in(0,1)\Rightarrow\xi^n\in(0,1)\Rightarrow 1-\xi^n\in(0,1)\Rightarrow a_n>0$ and $a_n\in(0,1)$
then
$$\begin{align}
a_{1}&=\xi^2(1-\xi)\\
a_{2}&=\sqrt{\xi^3(1-\xi)(1-\xi^2)}\\
&=\sqrt{\xi^3(1-\xi)^2(1+\xi)}\\
&=\xi(1-\xi)\sqrt{\xi(1+\xi)}
\end{align}$$
for $\xi\in(0,1)\Rightarrow \xi(1-\xi)>0$
$$\begin{align}
0<\xi&<1+\xi\\
0<\xi^2&<\xi(1+\xi)=\xi+\xi^2\\
\xi&<\sqrt{\xi(1+\xi)}\\
\xi^2(1-\xi)&<\xi(1-\xi)\sqrt{\xi(1+\xi)}\\
a_1&<a_2
\end{align}$$
then i think if i can proof that $a_{n+1}>a_n,n\in\mathbb{N}^*$ and $\xi\in(0,1)$ then the limit would be $1$

REPLY [2 votes]: I can't comment because apperently just guys with super high level can help you.
About this approach with the mean valued integral theorem: I've tried it.It won't work simply because $\xi$ depends on $n$ if it was a constant for any $n$ I would agree with you, this limit is 1. But the sequence $a_n$ you created is not necessarly monotonic. Because $\xi$ is a function of $n$