If $\sum _{r=1}^n{t}_r=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{8},$ where $t_r$ denotes the $r^{th}$ term of a series, then $\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{t_r}$ is

Given, $\sum _{r=1}^n{t}_r=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{8}=S_n$ (say)

$\therefore \sum _{r=1}^{n-1}{t}_r=\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{8}=S_{n-1}$

Now, $t_n=S_n-S_{n-1}=\frac{n\left(n+1\right)\left(n+2\right)}{2}$

$\therefore \underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{t_r}=\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{2}{n\left(n+1\right)\left(n+2\right)}$

$=\underset{n\to \infty }{\lim }\sum _{r=1}^n\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)$

$=-\underset{n\to \infty }{\lim }\sum _{r=1}^n\left(\frac{1}{\left(n+1\right)\left(n+2\right)}-\frac{1}{n\left(n+1\right)}\right)$

$=-\underset{n\to \infty }{\lim }\left(\frac{1}{\left(n+1\right)\left(n+2\right)}-\frac{1}{2}\right)$

$=-\left(0-\frac{1}{2}\right)=\frac{1}{2}$