Finding series from its sum, then finding its reciprocal's sum

Question

If $$\sum_{r=1}^nt_r=\frac{n(n+1)(n+2)}{12}$$, then value of $$\sum_{r=1}^n\frac{1}{t_r}$$ is

Now, how can we find the series (i.e. formulae of $n^{th}$ term) from the sum? The answer is $\frac{4n}{n+1}$

Answer 1

$$t_n=\frac{n(n+1)(n+2)}{4}-\frac{(n-1)n(n+1)}{4}=\frac{n(n+1)}{4}.$$ Now, use $$\frac{4}{n(n+1)}=4\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ and a telescopic summation.

Answer 2

Observe that $$ t_n=\frac{n(n+1)(n+2)}{12}-\frac{(n-1)n(n+1)}{12}=\frac{n(n+1)}{4} $$ and $$ \frac{1}{t_n}=\frac{4}{n(n+1)}=\frac{4}{n}-\frac{4}{n+1}, $$ and hence $$ \sum_{k=1}^n\frac{1}{t_k}=\left(\frac{4}{1}-\frac{4}{2}\right)+\left(\frac{4}{2}-\frac{4}{3}\right)+\cdots+ \left(\frac{4}{n-1}-\frac{4}{n}\right)+\left(\frac{4}{n}-\frac{4}{n+1}\right)\\ =\frac{4}{1}-\frac{4}{n+1}=\frac{4n}{n+1} $$

Answer 3

The sum and difference are inverse operations:

$\begin{align*} S_n &= \sum_{1 \le k\le n} t_k \\ t_n &= S_ n - S_{n - 1} \end{align*}$

Thus you can get the term $t_n$ knowing the sum, and set up the sum of reciprocals.