Show that $$ \begin{align} \left(\frac{\pi}{2}\right)^2\left[\int_0^{\pi/2}\cos^{2n}t\ dt-\int_0^{\pi/2}\cos^{2n+2}t\ dt\right]&=\frac{\pi^3}{8}\left[\frac{(2n-1)!!}{(2n)!!}-\frac{(2n+1)!!}{(2n+2)!!}\right]\\[10pt] &=\frac{\pi^3}{8}\frac{(2n-1)!!}{(2n+2)!!} \end{align} $$ Deduce that $$ 0<\frac{(2n)!!}{(2n-1)!!}I_{2n}\le\frac{\pi^3}{8}\frac{1}{2n+2} $$
I am working on an assignment which requires me to show that $$ \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6} $$ These are the last two parts that I cannot for the life of me figure out. If someone could expand it so I can understand it better that would be greatly appreciated. Thank you!
I'm assuming that you already show that the closed form of this Wallis' integral is $$\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}\frac{\pi}{2}\tag{1} $$ hence $$\frac{\pi^{2}}{4}\left(\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx-\int_{0}^{\pi/2}\cos^{2n+2}\left(x\right)dx\right) $$ $$=\frac{\pi^{3}}{8}\left(\frac{\left(2n-1\right)!!}{\left(2n\right)!!}-\frac{\left(2n+1\right)!!}{\left(2n+2\right)!!}\right)=\frac{\pi^{3}}{8}\left(\frac{\left(2n-1\right)!!\left(2n+2\right)-\left(2n+1\right)!!}{\left(2n+2\right)!!}\right) $$ $$=\frac{\pi^{3}}{8}\left(\frac{\left(2n-1\right)!!\left(2n+2-2n-1\right)}{\left(2n+2\right)!!}\right)=\color{red}{\frac{\pi^{3}}{8}\frac{\left(2n-1\right)!!}{\left(2n+2\right)!!}}\tag{2} $$ by the definition of the double factorial. For the second part note that, if $\left|x\right|\leq\frac{\pi}{2} $, we have $$x^{2}\leq\frac{\pi^{2}}{4}\sin^{2}\left(x\right) $$ so $$\frac{\pi^{2}}{4}\left(\int_{0}^{\pi/2}\cos^{2n}\left(x\right)dx-\int_{0}^{\pi/2}\cos^{2n+2}\left(x\right)dx\right)=\int_{0}^{\pi/2}\cos^{2n}\left(x\right)\frac{\pi^{2}}{4}\sin^{2}\left(x\right)dx\geq I_{2n} $$ hence from $(2)$
as wanted.