Let's Break Off This Sum First

${S_n} := \displaystyle \sum_{i=1}^n t_i$

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

${t_n^{-1}} = \frac {2}{n(n+1)(n+2)} = \frac {1}{n(n+1)} - \frac {1}{(n+1)(n+2)}$

$\displaystyle \lim_{n→\infty} \displaystyle \sum_{i=1}^n {t_i^{-1}} = \frac {1}{2} - \frac {1}{6} + \frac {1}{6} - \frac {1}{12} + \ldots$

$= \frac {1}{2}$

Note : The penultimate step involved telescoping series - sum , where all the terms except the first are canceled out in pairs.

This is not a solution but a hint .

We have the general formula for sum of n terms. Lets call it $\large S_{n}$

From $\large S_{n}$ , try to deduce the general expression for $\large n^{th}$ term, say $\large T_{n}$ . Remember, $\large T_{n}$ = $\large S_{n+1}$ - $\large S_{n}$ .

Since, we have to compute : $\displaystyle \lim_{n\to\infty} \sum_{r=1}^n \dfrac1{t_n}$ , substitute $\large t_{n}$ for the general expression you obtain & apply telescoping series.

The answer should evaluate to 0.5.

If you need more clarification on any step, do comment .

summation r (r+1)(r+2) = n (n+1)(n+2)(n+3)/4

this is a basic formula and can be easily derived

hence n (n+1)(n+2)(n+3)/8 = summation r(r+1)(r+2)/2

tr = r(r+1)(r+2)/2

1/tr = 2/r(r+1)(r+2) summation 1/tr = summation 2/r(r+1)(r+2) = sum (1/r - 1/(r+1)) - sum(1/(r+1) - 1/(r+2))

since n-> infinity

sum 1/tr = (1/1) - (1/infinity) - (1/(1+1)) - (1/infinity)

               =  1 - 0.5
               = 0.5