Finding $\lim\limits_{n\rightarrow \infty}\sum\limits_{r=1}^{n}\frac{1}{T_r}$ given $\sum\limits_{r=1}^{n}T_r=\frac{n(n+1)(n+2)(n+3)}{8}$ If $\displaystyle\sum_{r=1}^{n}T_r=\frac{n(n+1)(n+2)(n+3)}{8}$, then how can we find $\displaystyle\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{1}{T_r}$?

$$T_{n+1}=\sum_{r=1}^{n+1}T_r - \sum_{r=1}^{n}T_r = \frac{1}{2}(n+1)(n+2)(n+3)$$
So, $$\begin{align}
\sum_{n=1}^{m}\frac{1}{T_r} = \sum_{n=1}^{m}\frac{2}{n(n+1)(n+2)} 
&= \sum_{n=1}^{m}\frac{2(n+1)^2-2n(n+2)}{n(n+1)(n+2)} \\
&= \sum_{n=1}^{m}\frac{2(n+1)^2}{n(n+1)(n+2)} -\sum_{n=1}^{m}\frac{2n(n+2)}{n(n+1)(n+2)} \\
&= \sum_{n=1}^{m}\frac{1}{n}+\sum_{n=1}^{m}\frac{1}{n+2}-\sum_{n=1}^{m}\frac{2}{n+1}\end{align}$$
Can you take it from here?
