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 $\lim _{n\to \infty }\sum _{r=1}^{n}\left(\frac{1}{{T}_{r}}\right)$ is equal to :

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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\ma…

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