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Soroban Hello, Heirot!
The general term is: .$\displaystyle a_n \;=\;\dfrac{1}{n(n+1)(n+2)(n+3)}$
Partial fractions: .$\displaystyle \displaystyle \frac{1}{n(n+1)(n+2)(n+3)} \;=\;\frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} + \frac{D}{n+3}$
We find that: .$\displaystyle A = \frac{1}{6},\;B = -\frac{1}{2},\;C = \frac{1}{2},\;D = -\frac{1}{6}$
Hence: .$\displaystyle \displaystyle a_n \;=\;\frac{\frac{1}{6}}{n} - \frac{\frac{1}{2}}{n+1} + \frac{\frac{1}{2}}{n+2} - \frac{\frac{1}{6}}{n+3} \;=\;\frac{1}{6}\left(\frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$
Then: .$\displaystyle \displaystyle S \;=\;\frac{1}{6}\sum^{\infty}_{n=1}\left(\frac{1}{ n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right) $
Crank out several terms:
$\displaystyle \displaystyle S \;=\;\frac{1}{6}\bigg[\left(1 - \frac{3}{2} + \frac{3}{3} - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{3}{3} + \frac{3}{4} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6}\right) $
. . . . . . . $\displaystyle \displaystyle + \left(\frac{1}{4} - \frac{3}{5} + \frac{3}{6} - \frac{1}{7}\right) + \left(\frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8}\right) + \left(\frac{1}{6} - \frac{3}{7} + \frac{3}{8} - \frac{1}{9}\right)$
. . . . . . . . . $\displaystyle \displaystyle + \left(\frac{1}{7} - \frac{3}{8} + \frac{3}{9} - \frac{1}{10}\right) + \left(\frac{1}{8} - \frac{3}{9} + \frac{3}{10} - \frac{1}{11}\right) + \hdots \bigg]$
We find that nearly all the terms cancel out.
And we are left with: .$\displaystyle \displaystyle S \;=\;\frac{1}{6}\left(1 - \frac{3}{2} + \frac{1}{2} + \frac{1}{3}\right) \;=\;\frac{1}{18}$