If $ \displaystyle \sum_{r=1}^{n} T_r = \frac{(2n - 1)(2n + 1)(2n + 3)(2n + 5)}{64}$$, \: then\:\: $$ \displaystyle \lim_{n \to \infty} \displaystyle \sum_{r=1}^{n} \left( \frac{1}{T_r} \right)$ is equal to:
\( 0 \)
\( 1 \)
\( \frac{1}{3} \)
\( \frac{2}{3} \)