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I've tried several times but I don't know how to get the correct result. Can anyone please tell me what should I do with this sum? Thank you :)

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

Dragan Zrilić
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2 Answers2

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It is equivalent to $$ \sum_{n=1}^\infty \frac{(-1)^{n-1} (2n-1)}{2^{n-1}} = \left. \sum_{n=0}^\infty (2n+1) x^n \right|_{-1/2} $$ The above is the series for (can be shown using power series for rational functions) $$ \frac{x+1}{(x-1)^2} = \sum_{n=0}^\infty (2n+1) x^n $$

Henricus V.
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  • For the last equality the OP can use: http://math.stackexchange.com/questions/333192/solve-sum-nxn – Kal S. Apr 03 '16 at 18:20
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    @Test123 It can be solved by multiplying both sides by $(x-1)^2$ and setup a recurrence relation for the coefficients. – Henricus V. Apr 03 '16 at 18:20
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Note that $$\sum_{n=1}^\infty\frac{(-1)^{n-1}(2n-1)}{2^{n-1}}=2\sum_{n=1}^\infty\frac{(-1)^{n-1}n}{2^{n-1}}-\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2^{n-1}}$$

So, the question is, what is $\sum_{n=1}^\infty\frac{(-1)^{n-1}n}{2^{n-1}}$?

Well, note that in general, if $-1<x<1$, then $$\sum_{n=0}^{\infty}nx^{n-1}=1+2x+3x^2+4x^3+\dots=1+x+x^2+x^3+\dots +x(1+x+x^2+ \dots)+x^2(1+x+x^2+ \dots)+\dots=(1+x+x^2+\dots x^n)^2$$

S.C.B.
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  • Sorry but that didn't helped me much. Can you be more precise what should I do? – Dragan Zrilić Apr 03 '16 at 18:01
  • @DraganZrilić Well, inserting $x= -\frac{1}{2}$ gives us how to evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}n}{2^{n-1}}$, with some small error. – S.C.B. Apr 03 '16 at 18:02