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This is a homework question; I'm supposed to use power series to find the following sum: $$\sum_{n=1}^\infty \frac{n}{2^n}$$ I took the geometric series $$\frac{1}{1-x}=\sum_{n=0}^\infty {x^n}$$ and differentiated and multiplied both sides by x to get $$\frac{x}{(1-x)^2}=\sum_{n=1}^\infty {nx^n}$$ I'm stuck because I'm not sure how to make the $$\frac{1}{2^n}$$ term appear.

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Try $x=\frac 12$? –  Mark Bennet Mar 19 at 8:43
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As an aside, this type of infinite series is called a polylogarithm. –  Lucian Mar 19 at 9:17

1 Answer 1

For $x=\frac {1} {2}$, we have $\sum _{i} nx^n=\sum _{i} n(\frac {1} {2})^n=\sum _{i} \frac {n} {2^n}$

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