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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Sign up to join this communityI 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 $n \to \infty$.
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, $$ \begin{align} \int_0^\sqrt{6}\left(1-\frac{x^2}{6}+\frac{x^4}{120}\right)^n\;\mathrm{d}x &=\frac{1}{\sqrt{n}}\int_0^\sqrt{6n}\left(1-\frac{x^2}{6n}+\frac{x^4}{120n^2}\right)^n\;\mathrm{d}x\\ &=\frac{1}{\sqrt{n}}\int_0^\infty e^{-x^2/6}\left(1-\frac{x^4}{180n}-\frac{10x^6-x^8}{64800n^2}\right)\;\mathrm{d}x+\frac{1}{\sqrt{n}}O\left(\frac{1}{n^3}\right)\\ &=\sqrt{\frac{3\pi}{2n}}\left(1-\frac{3}{20n}+\frac{11}{160n^2}+O\left(\frac{1}{n^3}\right)\right)\tag{7} \end{align} $$