For $n\ge 0\,$ $I_n=\int_0^\frac{\pi}{2}\cos^{2n} x\,dx$ and $J_n=\int_0^\frac{\pi}{2}x^2\cos^{2n} x\,dx\,$ Find: $$ S=\lim\limits_{n\to\infty} \sum_{k=1}^n\left(\frac{J_{k-1}}{I_{k-1}}-\frac{J_{k}}{I_{k}}\right) $$
I am trying to compute this sum, so far I noticed that $S$ is a telescoping sum so it reduces to $$S=\lim\limits_{n\to\infty} \left(\frac{J_0}{I_0}-\frac{J_{n}}{I_{n}}\right)$$ is not so hard to evaluate the first part $J_0=\frac{\pi^3}{24}\,$ and $I_0=\frac{\pi}{2}$ this way $$S=\frac{\pi^2}{12}+\lim\limits_{n\to\infty} \frac{J_{n}}{I_{n}}$$ and I could show using beta function that $$I_n=\frac{1}{2}\beta(\frac{2n+1}{2};\frac{1}{2})=\frac{1}{2}\frac{\sqrt{\pi}\,\Gamma(\frac{2n+1}{2})}{n!}$$ I would love to get some help in order to evaluate $J_n$.
Since the sum telescopes, you don't have to calculate the integrals exactly, only estimate them. We have $(\cos{x})^{2n} = \exp{(2n\log{\cos{x}})}$, and $2\log{\cos{x}}$ has a single maximum at $0$, at which it $\sim -x^2$ so we can apply Laplace's method to both integrals to find the leading behaviour as $n \to \infty$. This gives $$ I_n = \int_0^{\pi/2} \cos^{2n}{x} \, dx \sim \int_0^{\infty} e^{-nx^2} \, dx = \frac{\sqrt{\pi}}{2\sqrt{n}}, $$ and $$ J_n = \int_0^{\pi/2} x^2 \cos^{2n}{x} \, dx \sim \int_0^{\infty} x^2 e^{-nx^2} \, dx = \frac{\sqrt{\pi}}{4n^{3/2}}, $$ so $J_n/I_n \sim 1/(2n) \to 0$.