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\[\sum _{n=1}^{\infty } \text{nx}^n\]

\[\Sigma_1^\infty x^n=\frac{1}{1-x}\] for -1

yes

taking derivatives we get \[\Sigma_1^\infty nx^{n-1}=\frac{1}{(1-x)^2}\]

okay

multiply both sides by x to get
\[\Sigma_1^\infty nx^n=\frac{x}{(1-x)^2}\]

Okay,thanks

but of course when n = 0 in your power series you get 0 anyway, so you might as well start at 1!