Finding the limiting value in the form of recurrence Now, this is actually a weird one, because I usually wouldn't suspect one of these to pop up on facebook; but is there anything to this problem, is it solvable, or is it just pure jitter, and if not, how is it solved?

$$\mathrm{For\space every\space integer\space} n \geq 0,$$
$$ I_n=\int_0^{\huge\frac{\pi}{2}}\cos^{2n}x\space\mathrm{d}x\space;\space J_n=\int^{\huge\frac{\pi}{2}}_0 x^2\cos^{2n}x\space \mathrm{d}x$$
$$\mathrm{Find\space the\space limit \space below:}$$
$$ \lim_{n \to + \infty} 2 \sum_{k=1}^n\left(\frac{J_{k-1}}{I_{k-1}}-\frac{J_k}{I_k}\right)$$

 A: Using integration by parts:
\begin{align}
I_n &= \int_0^\frac{\pi}{2}\cos^{2n}(x)\,dx\\
&= \int_0^\frac{\pi}{2}(x)'\cos^{2n}(x)\,dx\\
&= \underbrace{x\cos^{2n}x\,\Bigg\vert_0^\frac{\pi}{2}}_{=0} + \int_0^\frac{\pi}{2}2nx\cos^{(2n-1)}(x)\sin x\,dx\\
&= n\int_0^\frac{\pi}{2}(x^2)'\cos^{(2n-1)}(x)\sin x\,dx\\
&= \underbrace{nx^2\cos^{(2n-1)}(x)\sin x \,\Bigg\vert_0^\frac{\pi}{2}}_{=0}- n\int_0^\frac{\pi}{2}x^2\left(-(2n-1)\cos^{(2n-2)}(x)\sin x + \cos^{(2n-1)}(x)\cos x\right)\,dx\\ 
&= n(2n-1)\int_0^\frac{\pi}{2}x^2\cos^{(2n-2)}(x)\sin^2{x}\,dx -\int_0^\frac{\pi}{2}\cos^{2n}(x)\,dx\\
&= n(2n-1)\int_0^\frac{\pi}{2}x^2\cos^{(2n-2)}(x)(1-\cos^2x)\,dx -n\cdot I_n\\
\end{align} 
Thus:
$$(1+n)I_n = n(2n-1)(J_{n-1}-J_n)$$
$$I_n = \frac{n(2n-1)}{n+1}(J_{n-1}-J_n)$$
Notice that the sum is a telescoping one:
$$\sum_{k=1}^n\frac{J_{k-1}}{I_{k-1}} - \frac{J_{k}}{I_{k}} = \frac{J_0}{I_0} - \frac{J_n}{I_n}\xrightarrow{n\to\infty} \frac{J_0}{I_0} - \lim_{n\to\infty}\frac{J_n}{I_n} = \frac{\pi^2}{12}- \lim_{n\to\infty}\frac{J_n}{I_n}$$
where $$I_0 = \int_0^\frac{\pi}{2}dx = \frac{\pi}{2}$$
$$J_0 = \int_0^\frac{\pi}{2}x^2\,dx = \frac{\pi^3}{24}$$
We have:
$$\lim_{n\to\infty}\frac{J_n}{I_n} = \lim_{n\to\infty}\frac{J_n}{\frac{n(2n-1)}{n+1}(J_{n-1}-J_n)} = \lim_{n\to\infty}\frac{\frac{n+1}{n(2n-1)}}{\frac{J_{n-1}}{J_n}-1}$$
By plotting $\frac{J_{n-1}}{J_n}$ it seems that it converges to $0$.
If we assume that, we have $\lim_{n\to\infty}\frac{J_n}{I_n} = 0$ and finally:
$$2\sum_{k=1}^\infty\frac{J_{k-1}}{I_{k-1}} - \frac{J_{k}}{I_{k}} = \frac{\pi^2}{6}$$
A: I got with Mathematica
$I_n=\dfrac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right)}{2 \Gamma (n+1)}$
$J_n=\\=\dfrac{1}{\sqrt{\pi } n^3}4^{-n-1}\left(i \sqrt{\pi } \, _4F_3(-2 n,-n,-n,-n;1-n,1-n,1-n;-1)+4^n \Gamma (1-n) \Gamma \left(n+\frac{1}{2}\right) \left(i \left(\pi ^2 n^2-1\right)+i n^2 \psi ^{(1)}(n)+\pi ^2 n^2 \left(\cot (\pi  n)-i \csc ^2(\pi  n)\right)\right)\right.$
where $_4F_3$ is the generalized hypergeometric function
The series is almost intractable
$$2 \sum _{k=1}^{\infty } \left(\dfrac{j_{k-1}}{i_{k-1}}-\dfrac{j_k}{i_k}\right)$$
A: Hint:
By telescoping and assuming that the ratio of the integrals converges to $0$, we have $$S=2\frac{J_0}{I_0}=\frac23\left(\frac\pi2\right)^3\frac2\pi=\frac{\pi^2}6.$$
And the ratio converges to $0$ because for increasing $n$, you integrate $x^2$ and $1$ with a weight significantly nonzero on a smaller and smaller neighborhood of $0$.
A: Well, we all know it's a telescoping sum, so it's 
$$2\left(\frac{J_0}{I_0}-\lim_{n\to\infty}\frac{J_n}{I_n}\right).$$ Let's address the problem others evaded: a rigorous proof of 
$$\lim_{n\to\infty}\frac{J_n}{I_n}=0.$$
We know that $$\frac{1-\cos x}{x^2}=\frac12\,\left(\frac{\sin x/2}{x/2}\right)^2\ge\frac4{\pi^2}$$ for $x\in(0,\pi/2]$, since the function is monotone decreasing, and then, $$\frac{1-\cos^2 x}{x^2}=\frac{1-\cos x}{x^2}\left(1+\cos x\right)\ge\frac4{\pi^2},$$ too. This means
$$I_n-I_{n+1}\ge\frac4{\pi^2}\,J_n.$$ But it's easy to show by partial integration that $$\frac{I_{n+1}}{I_n}=\frac{2n+1}{2n+2},$$ and this means
$$\frac{J_n}{I_n}\le\frac{\pi^2}{8(n+1)}.$$
