Infinite Sums

This series can be written as

$\dfrac{1}{3}\left(\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{15} + ... \right) = \dfrac{1}{3} \displaystyle\sum_{k=2}^{\infty} \dfrac{2}{n(n + 1)} =$

$\dfrac{2}{3}\displaystyle\sum_{k=2}^{\infty}\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right) = \dfrac{2}{3}*\dfrac{1}{2} = \dfrac{1}{3},$

where the fact that the last summation was of a telescoping series was used.

Thus $m + n = 1 + 3 = \boxed{4}.$

Let $S = \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{15} + \dfrac{1}{21} + \cdots$ .

Now consider the following sum $X = \dfrac{1}{9} + \dfrac{1}{18} + \dfrac{1}{30} + \dfrac{1}{45} + \dfrac{1}{63} + \cdots = \dfrac{1}{3}S$ .

Taking a factor of $\dfrac{1}{3}$ out we get that, $X = \dfrac{1}{3}\left(\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{15} + \dfrac{1}{21} + \cdots\right)$ .

Now rewrite every denominator as such, $X = \dfrac{1}{3}\left(\dfrac{1}{3} + \dfrac{1}{2\times 3} + \dfrac{1}{2\times 5} + \dfrac{1}{3 \times 5} + \dfrac{1}{3 \times 7} + \dfrac{1}{4 \times 7} + \cdots\right)$ .

Notice that we can rewrite $X$ as $X = \dfrac{1}{3}\left(\dfrac{1}{3}\times\left(1 + \dfrac{1}{2}\right) + \dfrac{1}{5}\times\left(\dfrac{1}{2} + \dfrac{1}{3}\right) + \dfrac{1}{7}\times\left(\dfrac{1}{3} + \dfrac{1}{4}\right) + \cdots\right)$ .

Then $X = \dfrac{1}{3}\left(\dfrac{1}{3}\times\dfrac{3}{2} + \dfrac{1}{5}\times\dfrac{5}{6} + \dfrac{1}{7}\times\dfrac{7}{12} + \cdots\right)$ .

Now cancelling out, $X = \dfrac{1}{3}\left(\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12}+ \dfrac{1}{20} + \dfrac{1}{30} + \cdots\right)$ .

Finally taking out $\dfrac{1}{2}$ , we get that, $X = \dfrac{1}{3}\left(\dfrac{1}{2}\left(1 + \dfrac{1}{3} + \dfrac{1}{6}+ \dfrac{1}{10} + \dfrac{1}{15} + \dfrac{1}{21} + \cdots\right)\right)$ .

Notice now that, $X = \dfrac{1}{3}\left(\dfrac{1}{2}\left(1 + S\right)\right)$ and that therefore $\dfrac{1}{3}S = \dfrac{1}{3}\left(\dfrac{1}{2}\left(1 + S\right)\right)$ .

We get that, $S = \dfrac{1}{2}\left(1 + S\right)$ and $\therefore S = 1$ . But $X = \dfrac{1}{3}S = \dfrac{1}{3}$ so $m = 1, n = 3$ and $m + n = 1 + 3 = \boxed{4}$ .

sequence for denominator is 3/2(n^2+3n+2)