# 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 $$\frac{4n}{n+1}$$

• the index is $r$ but you have $t_n$? – user715522 Feb 25 at 10:50
• The posted solutions appear to assume that you wanted the initial sum to hold $\forall n$ but of course you don't say that anywhere. If you are making that assumption, you should make it explicitly. – lulu Feb 25 at 10:54

$$t_n=\frac{n(n+1)(n+2)}{4}-\frac{(n-1)n(n+1)}{4}=\frac{n(n+1)}{4}.$$ Now, use $$\frac{4}{n(n+1)}=4\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ and a telescopic summation.
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 $$\sum_{k=1}^n\frac{1}{t_k}=\left(\frac{4}{1}-\frac{4}{2}\right)+\left(\frac{4}{2}-\frac{4}{3}\right)+\cdots+ \left(\frac{4}{n-1}-\frac{4}{n}\right)+\left(\frac{4}{n}-\frac{4}{n+1}\right)\\ =\frac{4}{1}-\frac{4}{n+1}=\frac{4n}{n+1}$$
\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.