User talk:75.15.160.108

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\sum_{i=1}^\infty i

\sum_{i=1}^\infty i(-1)^{i+1}

0.999...=0.9+0.09+0.009+...=\sum_{i=1}^\infty 0.9(0.1)^i

S=\frac{a_1}{1-r}

S=\frac{0.9}{1-0.1}=\frac{0.9}{0.9}=1

\zeta(s) = \sum_{n=1}^\infty {n^{-s}}