If \( \sum_{r=1}^{n} T_{r}=\frac{n(n+1)(n+2)(n+3)}{8} \), then \( \lim _{n \rightarrow \infty} \...
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If \( \sum_{r=1}^{n} T_{r}=\frac{n(n+1)(n+2)(n+3)}{8} \), then \( \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{T_{r}}= \)
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