Let $S_n=\sum_{r=1}^n T_r=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}.$ We need $\lim_{n\to\infty}\sum_{r=1}^n \frac{1}{T_r}.$

First find $T_n$ from the relation $T_n=S_n-S_{n-1}.$

Now, $S_n=\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$ and $S_{n-1}=\frac{(2n-3)(2n-1)(2n+1)(2n+3)}{64}.$

Therefore, \begin{align*} T_n&=\frac{(2n-1)(2n+1)(2n+3)}{64}\Big[(2n+5)-(2n-3)\Big]\ &=\frac{(2n-1)(2n+1)(2n+3)}{64}\cdot 8\ &=\frac{(2n-1)(2n+1)(2n+3)}{8}. \end{align*}

Hence $\frac{1}{T_r}=\frac{8}{(2r-1)(2r+1)(2r+3)}.$

We now decompose it into telescoping form. Observe that

Now sum from $r=1$ to $n$: \begin{align*} \sum_{r=1}^n \frac{1}{T_r} &=2\sum_{r=1}^n\left(\frac{1}{(2r-1)(2r+1)}-\frac{1}{(2r+1)(2r+3)}\right). \end{align*}

This is telescoping: \begin{align*} \sum_{r=1}^n \frac{1}{T_r} &=2\left(\frac{1}{1\cdot 3}-\frac{1}{3\cdot 5}+\frac{1}{3\cdot 5}-\frac{1}{5\cdot 7}+\cdots+\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)}\right)\ &=2\left(\frac{1}{3}-\frac{1}{(2n+1)(2n+3)}\right). \end{align*}

Taking limit as $n\to\infty$:

Hence the correct option is $\boxed{\text{A }\frac{2}{3}}.$

Comparison with stored answer: Stored correct answer is A, which matches our result.