Infinitely many capacitors

Possible Solution to this problem:

First of all, I wanted to determine the roots of the equation: $\sin\sqrt{x}=0$

And we have:
$\sin\sqrt{x}=0 , \sqrt{x} \in \left \{ \sin^{-1}(0) + k\pi | k\in \mathbb{Z} \right \} , \sqrt{x} \in \left \{ k\pi | k\in \mathbb{Z} \right \}$
So we get:
$x \in \left \{ (k\pi)^2 | k\in \mathbb{Z} \right \}$ from this we get: $r{_{k}} = (k\pi)^2,k \in \mathbb{Z}$
Replacing k with 1,2,...,n,n+1 we get: $r{_{0}}=0,r{_{1}}=\pi^2,r{_{2}}=(2\pi)^2),...,r{_{n}}=(n\pi)^2,r{_{n+1}}=[(n+1)\pi]^2$

The second step I did is resolving the integral after I determined: $r{_{n}},r{_{n+1}}$ :

$\int_{r_{n}}^{r_{n+1}}{f(x)dx} = \int_{(n\pi)^2}^{[(n+1)\pi]^2}\sin\sqrt{x}dx = 2\int_{(n\pi)^2}^{[(n+1)\pi]^2}\sqrt{x}\frac{\sin\sqrt{x}}{2\sqrt{x}}dx$ Now we will make the following substitution:
$\sqrt{x}=t , \frac{1}{2\sqrt{x}}dx = dt$ for $x=(n\pi)^2, t=n\pi$ and for $x=[(n+1)\pi]^2, t=(n+1)\pi$ and we get:
$2\int_{n\pi}^{(n+1)\pi}t\sin(t)dt=2\int_{n\pi}^{(n+1)\pi}t(-\cos(t))'dt=-2t\cos(t) |_{n\pi}^{(n+1)\pi}+2\int_{n\pi}^{(n+1)\pi}\cos(t)dt=$
$=-2(n+1)\pi \cos((n+1)\pi)+2n\pi \cos(n\pi)+2\sin(t)|_{n\pi}^{(n+1)\pi}=2\pi(n\cos(n\pi)-(n+1)\pi\cos[(n+1)\pi])$

We also know that: $cos(n\pi+\pi)=\cos(n\pi)cos\pi-\sin(n\pi)\sin\pi=-\cos(n\pi)$
We get that: $2\pi(n\cos(n\pi)-(n+1)\pi\cos[(n+1)\pi])=2\pi(n\cos(n\pi)+(n+1)\cos(n\pi))=$
$=2\pi(2n\cos(n\pi)+\cos(n\pi))=2\pi(2n+1)cos(n\pi)$

After getting this, we keep in mind that: $\cos(n\pi)=(-1)^n$

Now we go back to the sum:

$\sum_{n=0}^{\infty}( \int_{r_{n}}^{r_{n+1}}f(x)dx )^{-1}=\sum_{n=0}^{\infty}(2\pi(2n+1)(-1)^n)^{-1}=\frac{1}{2\pi} \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}$

And now the fun part:

Consider the sequence: $S(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}=S'(x)=\sum_{n=0}^{\infty}(-1)^nx^{2n}=1-x^2+x^4-...=\frac{1}{1+x^2}$ And so: $S(x)=\int \frac{1}{1+x^2}dx = \arctan{x}+K$ $S(0)=0 , \arctan{0} + K=0, K=0$

Going back to our first sum we replace x with 1, $x = 1$ and we get that: $\frac{1}{2\pi}\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{1}{2\pi}\arctan(1)=\frac{1}{2\pi}\frac{\pi}{4}=\frac{1}{8}$

Finally: $\frac{1}{C}=\frac{1}{8} => C=8$