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I have problem with showing that the limit of the following function

$$\frac{ \sqrt{\frac{3 \pi}{2n}} - \int_0^{\sqrt 6}( 1-\frac{x^2}{6} +\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$

equal to $1$, with $n \to \infty$.

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For what it counts, Wolphram alpha gives… $\int_0^{\sqrt 6}(1-\frac{x^2}{6}+\frac{x^4}{120})^ndx= \sqrt6 F_1(\frac{1}{2};-n,-n; -\frac{3}{2};\frac{3i}{5i+\sqrt5},-\frac{3i}{-5i+\sqrt5})$ Where $F_1(a;b_1,b_2;c;x,y)$ is Appell Hypergeometric function of two variables. – NikBels Jan 4 '12 at 21:25
I thought that the definite integral could be treated as a constant but Nikhil Bellarykar clarified that it would be a function of n, so using L'Hospital rule is not possible. – NoChance Jan 4 '12 at 21:34
Wolphram alpha further gives the series expansion of the Appell function at $n=\infty$ as ${(\frac{3}{10})^n}(\frac{64n^2+56n}{27}+1)$.… But if we plug in this value in the original function, the answer does not come as 1. Am I wrong to simply plug in this value? maybe. Even if I just try to plug in a simple series expansion, the result does not seem to be 1. Someone please clarify what is missing from my approach. – NikBels Jan 4 '12 at 21:55
up vote 6 down vote accepted

Hint: First note that for $x\in[0,\sqrt{6}]$, $1-\frac{x^2}{6}+\frac{x^4}{120}$ monotonically decreases from $1$ to $\frac{3}{10}$ and that $$ 1-\frac{x^2}{6}+\frac{x^4}{120}\le1-\frac{x^2}{9}\le e^{-x^2/9}\tag{1} $$ You might try the change of variables $x\mapsto x/\sqrt{n}$ so that $$ \int_0^\sqrt{6}\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)^n\;\mathrm{d}x =\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x\tag{2} $$ and $(1)$ then says that for $x\in[0,\sqrt{6n}]$ $$ \left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\le e^{-x^2/9}\tag{3} $$ Thus, we have $$ \begin{align} \int_a^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x &\le\int_a^\infty e^{-x^2/9}\;\mathrm{d}x\\ &\le\frac{9}{2a}e^{-a^2/9}\tag{4} \end{align} $$ Notice that the integrand on the right in $(2)$ tends to $e^{-x^2/6}$, so consider $$ \begin{align} \left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n &=e^{-x^2/6}\exp\left(n\log\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)+\frac{x^2}{6}\right)\\ &=e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}+O\left(\frac{x^8}{n^3}\right)\right)\tag{5} \end{align} $$ Applying $(5)$ to $(2)$ yields $$ \begin{align} &\left|\;\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x -\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right)\;\mathrm{d}x\;\right|\\ &\le\frac{1}{\sqrt{n}}\int_0^{\log(n)} e^{-x^2/6}O\left(\frac{x^8}{n^3}\right)\;\mathrm{d}x\\ &+\frac{1}{\sqrt{n}}\int_{\log(n)}^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x +\frac{1}{\sqrt{n}}\int_{\log(n)}^\infty e^{-x^2/6}\left|1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right|\;\mathrm{d}x\\ &=\frac{1}{\sqrt{n}}O\left(\frac{1}{n^3}\right)\tag{6} \end{align} $$ Therefore, $$ \begin{align} \int_0^\sqrt{6}\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)^n\;\mathrm{d}x &=\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x\\ &=\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right)\;\mathrm{d}x+\frac{1}{\sqrt{n}}O\left(\frac{1}{n^3}\right)\\ &=\sqrt{\frac{3\pi}{2n}}\left(1-\frac{3}{20n}+\frac{11}{160n^2}+O\left(\frac{1}{n^3}\right)\right)\tag{7} \end{align} $$

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Truly amazing effort! – NoChance Jan 7 '12 at 11:09
Thank you very much, robjohn. – David Jan 7 '12 at 14:20
@David: I was really hoping that there was a simpler solution. If the instructor shows you a simpler solution, please post it here. – robjohn Jan 8 '12 at 0:35
Sure. Thank you very much, robjohn. – David Jan 8 '12 at 0:46
@robjohn: you have interesting proof. But how to show that there are no other terms with $1/n$, $1/n^2$ so on? – Aleks.K Jun 18 '12 at 19:41

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