Calculate $\sum^{\infty}_{r=1} \frac{8}{(2r-1)\sqrt{(2r+3)(2r+5)}+\sqrt{(2r+1)(2r-1)}(2r+3)}$ If $\sum^{\infty}_{r=1} \frac{8}{(2r-1)\sqrt{(2r+3)(2r+5)}+\sqrt{(2r+1)(2r-1)}(2r+3)}=\sqrt{a}+\sqrt{\frac{5}{3}}-\sqrt{c}$, where $a$ and $c$ are coprime numbers, then the value of $a+\frac{c}{4}$ is
(1) 3
(2) 5
(3) 8
(4) 4
I'm completely lost here, some pointers would be nice.
I tried putting it into Wolfram Alpha to see if it could get a partial sum formula, but that didn't seem to work.
And putting in the entire series itself just gave me a decimal approximation (unless I'm missing something).

Hint: Use telescoping series. Multiply numerator and denominator by $\sqrt{\frac{2r+5}{2r+3}}-\sqrt{\frac{2r+1}{2r-1}}$, to get final simplified expression for $t_{r}$ as under.
$$\begin{aligned}t_{r}&=\frac{8}{(2r-1)(2r+3)\left[\sqrt{\frac{2r+5}{2r+3}}+\sqrt{\frac{2r+1}{2r-1}}\right]} \\ &= \sqrt{\frac{2r+1}{2r-1}}-\sqrt{\frac{2r+5}{2r+3}}\end{aligned}$$
Now define $a_{r}=\sqrt{\frac{2r+1}{2r-1}}$, so that $t_{r}=a_{r}-a_{r+2}$. The sum would be $\sum_{r=1}^{\infty}(a_{r}-a_{r+2})=a_{1}+a_{2}-2a_{\infty}$. Since $a_{\infty}\to 1$, $a_{1}=\sqrt{3}$ and $a_{2}=\sqrt{5/3}$. So the answer seems to be $\sqrt{3}+\sqrt{\frac{5}{3}}-\sqrt{4}\approx 1.01808$.
