My first thought is that each term is of the form 1/ ( n * (n + 18) ) . Represent term as ( A/n - B/(n+18)), compute partial sums using that form and check for telescoping cancellation.
It appears that the sequence \(\displaystyle 2,20,38,56,\cdots\) is \(\displaystyle a_n=18\cdot n-16\).
Thus \(\displaystyle \sum\limits_{k = 1}^\infty {\frac{1}{{(18k - 16)(18k + 2)}}} \) has a sum SEE HERE
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