If \(\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{~T}_{\mathrm{r}}=\frac{(2 \mathrm{n}-1)(2 \mathrm{n}+1)(2 \mathrm{n}+3)(2 \mathrm{n}+5)}{64}\), then \(\lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{r}=1}^{\mathrm{n}}\left(\frac{1}{T_{\mathrm{r}}}\right)\) is equal to :
1
1
2
0
3
\(\frac{2}{3}\)
4
\(\frac{1}{3}\)