Finding series from its sum, then finding its reciprocal's sum 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 $\frac{4n}{n+1}$

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
$$
\sum_{k=1}^n\frac{1}{t_k}=\left(\frac{4}{1}-\frac{4}{2}\right)+\left(\frac{4}{2}-\frac{4}{3}\right)+\cdots+
\left(\frac{4}{n-1}-\frac{4}{n}\right)+\left(\frac{4}{n}-\frac{4}{n+1}\right)\\ =\frac{4}{1}-\frac{4}{n+1}=\frac{4n}{n+1}
$$
