Summation and Summation

$\displaystyle f(s)=\sum_{n=0}^\infty\frac{1}{s^n}=\cfrac{1}{1-\cfrac{1}{s}}=\frac{s}{s-1}\implies \sum_{s=2}^{10}f(s)=\sum_{s=2}^{10}\frac{s}{s-1}=\frac{2}{1}+\frac{3}{2}+\frac{4}{3}+\cdots+\frac{10}{9}=\frac{29809}{2520}\implies \text{answer}=29809+2520=\boxed{32329}$

A slightly more algebraic solution.

$\displaystyle \sum _{s=2} ^{10} {f(s)} = \sum _{s=2} ^{10} {\sum _{n=0} ^\infty \frac{1}{s^n} } = \sum _{s=2} ^{10} {\frac {1}{1-\frac{1}{s}}} = \sum _{s=2} ^{10} {\frac {s}{s-1}} = \sum _{s=1} ^{9} {\frac {s+1}{s}}$

$\displaystyle = \sum _{s=1} ^{9} {\left( 1 + \frac {1}{s} \right)} = \sum _{s=1} ^{9} {1 } + \sum _{s=1} ^{9} {\frac {1}{s}}$

$\displaystyle = 9 + \left( \frac {1}{1} + \frac {1}{2} + \frac {1}{4} + \frac {1}{8} \right) + \left( \frac {1}{3} + \frac {1}{6} + \frac {1}{9} \right) + \frac {1}{5} + \frac {1}{7}$

$\displaystyle = 9 + \frac {1-\left( \frac{1}{2}\right)^4} {1-\frac{1}{2}} + \frac {1}{3} \left( \frac {1}{1} + \frac {1}{2} + \frac {1}{3} \right) + \frac {1}{5} + \frac {1}{7}$

$\displaystyle = 9 + \frac {15}{8} + \frac {11}{18} + \frac {12}{35} = \frac {22680+4725+1540+864} {2520} = \frac {29809} {2520} = \frac {A}{B}$

$\Rightarrow A+B = 29809+2520= \boxed{32329}$

Large calculation !!!!!!!