# how to compute this limit

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 $1$

• Very Nice question! – Peter Jan 28 '15 at 17:30
• the Product $(1-x)....(1-x^n)$ is known as a PochhammerQ Symbol which have some interesting asymptotic expansions... – tired Jan 28 '15 at 17:34
• We have: $$I_n = \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \geq \int_{0}^{1}x^{n+1}(1-x)^n\,dx = \frac{1}{(n+1)\binom{2n+2}{n+1}},$$ so the limit is $\geq \frac{1}{4}$, but I am having a hard time in proving a converse bound in order to show that the limit is exactly $\frac{1}{4}$. By a simple argument, the limit is $\leq 1$. – Jack D'Aurizio Jan 28 '15 at 17:39
• In fact, the limit seems to be $1$. – Peter Jan 28 '15 at 17:43
• Perhaps a shot in the dark, but parameterizing the integral as $$I_{a,n}=\int_0^1 (ax)^{n+1}(1-x)\cdots(1-x^n)\,dx$$ and differentiating $n+1$ times gives the differential equation ${I_a}^{(n+1)}-(n+1)!I_a=0$ with characteristic equation $r^{n+1}-(n+1)!=0$. – user170231 Jan 28 '15 at 17:47

We have: $$\begin{eqnarray*} I_n &=& \int_{0}^{1}x^{n+1}(1-x)(1-x^2)\cdot\ldots\cdot(1-x^n)\,dx \\&\leq& \int_{0}^{1}x^{n+1}(1-x^n)^n\,dx = \frac{\Gamma(n)\,\Gamma\left(1+\frac{2}{n}\right)}{\Gamma\left(2+n+\frac{2}{n}\right)}\leq\frac{1}{n^2},\tag{1}\end{eqnarray*}$$ while on the other hand, if we set: $$A(n,m)\triangleq\int_{0}^{1} x^n \prod_{k=1}^{m}(1-x^k)\,dx,\qquad B(n,m)=\int_{0}^{1} x^n (1-x^m)^m\,dx \tag{2}$$ we have: $$\begin{eqnarray*} I_n = A(n+1,n) &=& A(n+1,n-1)-A(2n+1,n-1)\\&=&A(n+1,n-2)- A(2n,n-2)-A(2n+1,n-1)\end{eqnarray*}$$ and since $A(2n+2-i,n-i)\leq B(2n+2-i,n-i)\leq\frac{1}{(n+2)^3}$ it follows that: $$I_n = A(n+1,1)-\sum_{k=1}^{n-1}A(2n+2-k,n-k) \geq \frac{1}{(n+2)(n+3)}-\frac{n-1}{(n+2)^3}\tag{3}$$ so $I_n\geq\frac{2}{(n+2)^3}$ and this inequality, together with $(1)$, implies: $$I_n^{\frac{1}{n}}\xrightarrow[n\to +\infty]{}\color{red}{1}. \tag{4}$$
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 $n$