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