use the lim(n->∞)(An+1/An) rule
A(n+1)/A(n) = [(n+1)^2012 0.95^(n+1)] / [ n^2012 0.95^n ] = ( 1 + (1/n) )^2012 0.95 -> 0.95 < 1 so the series converges
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A(n+1)/A(n) = [(n+1)^2012 0.95^(n+1)] / [ n^2012 0.95^n ] = ( 1 + (1/n) )^2012 0.95 -> 0.95 < 1 so the series converges