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Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$. Is this limit equal to $\lim\limits_{n{\rightarrow}\infty}(a_{n})^{\frac{1}{2}}$? I am unsure if I can apply l'Hospital to this limit, as the integral is a real value. I tried applying integration by parts but I ended nowhere. Later edit: I am so sorry, it is $(1-x)^{n}$, not $(1-x^{n})$, I did this mistake out of tiredness.

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    $\begingroup$ You can evaluate $a_n$ explicitly. $\endgroup$ – i707107 Mar 3 at 20:02
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    $\begingroup$ did you mean $a_{n}=\int_{0}^{1} x^{n}(1-x)^n dx$ ? because it's easy to show $\int_{0}^{1} x^{n}(1-x^{n})dx$ $\endgroup$ – LAGRIDA Mar 3 at 20:04
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    $\begingroup$ Isn't it $a_n=\int_0^1x^ndx-\int_0^1x^{2n}dx=\frac{1}{n+1}-\frac{1}{2n+1}=\frac{n}{(n+1)(2n+1)}$? $\endgroup$ – JustDroppedIn Mar 3 at 20:04
  • $\begingroup$ Ooops. It's $(1-x)^{n}$ $\endgroup$ – Septimiu Cristian Mar 4 at 1:26
7
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Starting with integration by parts we have

$$\begin{aligned} \displaystyle a_n \displaystyle & = \int_0^1 x^n(1-x)^n\,{dx} \\& = \frac{1}{n+1}\int_0^1 (x^{n+1})'(1-x)^n\,{dx} \\& = \frac{x^{n+1}}{n+1}(1-x)^n\,\bigg|_0^1 -\frac{1}{n+1}\int_0^1 x^{n+1}[(1-x)^n]'\,{dx} \\& = \frac{n}{n+1}\int_0^1 x^{n+1}(1-x)^{n-1}\,{dx} \end{aligned} $$

But also by symmetry this is the same as $\displaystyle a_n = \frac{n}{n+1}\int_0^1 x^{n-1}(1-x)^{n+1}\,{dx}$, therefore

$$\begin{aligned} \displaystyle 2a_{n} & = \frac{n}{n+1}\int_0^1 x^{n+1}(1-x)^{n-1}\,d{x}+ \frac{n}{n+1}\int_0^1 x^{n-1}(1-x)^{n+1}\,{dx} \\& = \frac{n}{n+1}\int_0^1 (2x^2-2x+1)x^{n-1}(1-x)^{n-1}\,{dx} \\& = -\frac{2n}{n+1} a_n+\frac{n}{n+1}a_{n-1}\end{aligned}$$

Therefore $\displaystyle a_{n} = \frac{na_{n-1}}{4n+2}$, hence $\displaystyle a_{n+1} = \frac{(n+1)a_{n}}{4n+6}$ so $\displaystyle \frac{a_{n+1}}{a_n} = \frac{(n+1)}{4n+6} \to \frac{1}{4}. $

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    $\begingroup$ Nice solution, and thanks for the catch! $\endgroup$ – Isaac Browne Mar 4 at 1:02
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    $\begingroup$ Nice use of symmetry. $\endgroup$ – Claude Leibovici Mar 4 at 8:34
4
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If the integrand is $x^n (1-x^n)$, you can integrate directly, as stated in the comments. But if it is $x^n (1-x)^n$, then this is the Beta Function, and we can rewrite it in terms of the gamma function as $$a_n = \frac{\Gamma (n+1) \Gamma(n+1)}{\Gamma(2n+2)}$$ And so $$a_{n}/a_{n+1} = \frac{\Gamma(n+1)^2}{\Gamma(n+2)^2} \cdot \frac{\Gamma(2n+4)}{\Gamma(2n+2)} = \frac{1}{(n+1)^2} \cdot \frac{(2n+3)(2n+2)}{1}$$ And the limit of this is simply 4.

This would not be equal to the limit of $a_n^{1/2}$, as that would just be $0$, the same limit as $a_n$.

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You are considering $$ a_n = \int_0^1 x^n(1-x)^n\,{dx} $$ Let $x=\sin^2(t)$ to make $$a_n=2 \int_0^{\frac \pi 2}\sin ^{2 n+1}(t) \cos ^{2 n+1}(t)\,dt=B(n+1,n+1)=\frac{\sqrt{\pi } \,\, \Gamma (n+1)}{2^{2 n+1}\Gamma \left(n+\frac{3}{2}\right)}$$ from which follows NoName's result for ${a_{n + 1} \over a_{n}}$.

Interesting would be to take the logarithm of $a_n$, to use Stirling approximation for the gamma function and continuing with Taylor expansion to get $$\log(a_n)=-2 n \log (2)+\left(\frac{1}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{\sqrt{\pi }}{2}\right)\right)-\frac{3}{8 n}+O\left(\frac{1}{n^2}\right)$$ which makes $$a_n \sim 2^{-(2 n+1)} \sqrt{\frac{\pi}{n}}\, \left(1-\frac{3}{8 n}+\frac{25}{128 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ which is quite accurate.

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Lemma Suppose that $f$ is a continuous positive function, then $\lim_{n \to \infty} \frac{ \int_0^1 f^{n+1}dx }{ \int_0^1 f^{n} dx} = \max f$

So, using the Lemma we conclude that

$$ \lim _{n\to \infty} \frac{\int _0^1x^{n+1}(1-x)^{n+1} dx}{\int_0 ^1x^n(1-x)^n dx }=\max_{x \in [0,1] } x(1-x)=\frac{1}{4}$$

Proof of the Lemma: We may assume that $\max f=1$. For the upper bound notice \begin{align} \frac{ \int_0^1 f^{n+1}dx }{ \int_0^1f^{n} dx}&= \frac{ \int_0^1 f \cdot f^{n} fdx }{ \int_0^1f^{n} dx}\\ \leq 1 \end{align}

Define $\mathcal{N}_{\epsilon}= \{x| f>1-\epsilon \}$. We provide now the lower bound,

\begin{align} \frac{ \int_0^1 f^{n+1}dx }{ \int_0^1 f^{n} dx}&\geq \frac{ \int_{\mathcal{N}_{\epsilon}} \cdot f^{n+1} fdx }{ \int f^{n} dx}\\ &\geq (1-\epsilon) \frac{ \int_{\mathcal{N}_{\epsilon}} \cdot f^{n} fdx }{ \int f^{n} dx} \to 1-\epsilon \end{align}

For the last limit observe the following \begin{align} \frac{\int_{[0,1]\setminus \mathcal{N}_{\epsilon}} f^n} {\int_0^1 f^n}&\leq \frac{|[0,1]\setminus \mathcal{N}_{\epsilon}| (1-\epsilon)^n}{\int_{\mathcal{N}_{\epsilon/2}} f^n} \\ &\frac{|[0,1]\setminus \mathcal{N}_{\epsilon}| (1-\epsilon)^n}{|\mathcal{N}_{\epsilon/2}| (1-\epsilon/2)^n} \\ &C \left ( \frac{1-\epsilon}{1-\epsilon/2}\right )^n \to 0 \end{align}

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}x^{n}\pars{1 - x}^{n}\dd x} = \int_{-1/2}^{1/2}\pars{{1 \over 4} - x^{2}}^{n}\dd x = {1 \over 2^{2n}}\int_{0}^{1}\pars{1 - x^{2}}^{n}\dd x \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,& {1 \over 2^{2n}}\int_{0}^{\infty}\expo{-nx^{2}}\dd x = {\root{\pi} \over 2^{2n + 1}\root{n}} \end{align} such that $$ {a_{n + 1} \over a_{n}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, {1/4 \over \root{1 + 1/n}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,\bbx{1 \over 4} $$

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