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This question already has an answer here:

It is obvious that

$\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}(2n-1)}{2^{n-1}}$

is alternating series, therefore, before we could talk about finding the sum of the series, we have to check if it's unconditionally convergent, if we look at $a_n=\frac{2n-1}{2^{n-1}}$ it is obvious that we could use d'Alembert's test and easily prove that this is unconditionally convergent series. So now, we should find the sum:

$S_n=\sum_{k=1}^{\infty} \frac{(-1)^{n-1}(2n-1)}{2^{n-1}}$

because of the denominator, it really looks like geometric series, but i could't find a way to simplify it in order to determine the sum. Any ideas?

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marked as duplicate by user296602, Crostul, Kamil Jarosz, Adam Hughes, user147263 Apr 2 '16 at 23:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Split it into $$2 \sum_{n \ge 0} n \left( - \frac{1}{2} \right)^n + \sum_{n \ge 0} \left( - \frac{1}{2} \right)^n$$ $\endgroup$ – Crostul Apr 2 '16 at 16:30
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define $A_n(x)=\sum_{n=1}^{+\infty} x^n \; \; |x|<1$ then we have: $$A_n(x)=\sum_{n=0}^{+\infty} x^n=\frac{1}{1-x}$$ now differentiate with respect to $x$ and get: $$A_n'(x)=\sum_{n=1}^{+\infty} nx^{n-1}=\frac{1}{(1-x)^2}$$ so by letting $x=\frac{-1}{2}$ we get: $$A_n(\frac{-1}{2})=\sum_{n=0}^{+\infty} (\frac{-1}{2})^n=\frac{2}{3} \Rightarrow \sum_{n=1}^{+\infty} (\frac{-1}{2})^n=\frac{-1}{3} \Rightarrow \sum_{n=1}^{+\infty} (\frac{(-1)^n}{2^{n-1}})=\frac{-2}{3}$$ and also: $$A_n'(\frac{-1}{2})=\sum_{n=1}^{+\infty} n(\frac{-1}{2})^{n-1}=\frac{4}{9} \Rightarrow \sum_{n=1}^{+\infty} 2n(\frac{-1}{2})^{n-1}=\frac{8}{9}$$ sum up these two and you get the answer: $$S_n=\frac{2}{9}=\sum_{n=1}^{+\infty} 2n(\frac{-1}{2})^{n-1}+\sum_{n=1}^{+\infty} (\frac{(-1)^n}{2^{n-1}})$$

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I will try to generalize your question a little bit. Using differentiation of power series, provided that it converges $\sum\limits_{n=1}^\infty P(n)a^{n}$ (where P is a polynomial) can be calculated differentiating the equality $\sum\limits_{n=0}^\infty x^n$=$\frac{1}{1-x}$ , shifting indices then evaluate at the value you want .

I think, it is good to have in mind that "polynomial times geometric is no harder than to sum a geometric series".

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