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I`m trying to evaluate this series and would like to get some advice how to do that.
What I need to find here to start with? $$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}= \frac{1}{2} + \frac{3}{4}+\frac{5}{8}+\dots$$

jimjim
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Ofir Attia
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4 Answers4

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Note that $$\dfrac{2n-1}{2^n} = \dfrac{n}{2^{n-1}} - \dfrac1{2^n}$$

Now make use of the following. For $\vert x \vert < 1$, we have $$\sum_{n=0}^{\infty} x^n = \dfrac1{1-x}$$ Differentiating both sides, we get that $$\sum_{n=0}^{\infty} nx^{n-1} = \dfrac1{(1-x)^2}$$ Take $x=1/2$ to get what you want.

  • Also don't forget that these formulas work when $n \ge 0$, but the series in the question has $n \ge 1$. – Javier Jul 01 '13 at 19:53
  • That was meant for the asker, by the way. – Javier Jul 01 '13 at 19:54
  • How did you get to $x^n$? this is what I didnt understand. – Ofir Attia Jul 01 '13 at 20:49
  • @Offir Attia, The reason the are using $x^n$ is that your series has

    $$\frac{1}{2^n} = \left( \frac{1}{2} \right)^n.$$

    This is the geometric series which is what they are using to solve the problem. You should memorize the result for the geometric series and look for it in the problems you solve.

     – Spencer Jul 01 '13 at 21:08
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Hints:

$$\begin{align*}\bullet&\;\;\;|x|<1\implies \sum_{n=0}^\infty x^n=\frac1{1-x}\;,\;\;\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}\\ \bullet&\;\;\;\frac{2n-1}{2^n}=n\frac1{2^{n-1}}-\frac1{2^n}\end{align*}$$

DonAntonio
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$$ \displaylines{ s = \sum\limits_{n = 1}^{ + \infty } {\frac{{2n - 1}}{{2^n }}} = \sum\limits_{n = 0}^{ + \infty } {\frac{{2n + 1}}{{2^{n + 1} }}} = \frac{1}{2}\sum\limits_{n = 0}^{ + \infty } {\frac{{2n + 1}}{{2^n }}} \cr \Leftrightarrow s = \frac{1}{2} + \frac{1}{2}\sum\limits_{n = 1}^{ + \infty } {\frac{{2n - 1}}{{2^n }}} + \frac{1}{2}\sum\limits_{n = 0}^{ + \infty } {\frac{2}{{2^{n + 1} }}} \cr \Leftrightarrow s - \frac{1}{2}s = \frac{1}{2} + \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{2^{n + 1} }}} \cr \Leftrightarrow \frac{1}{2}s = \frac{1}{2} + \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{2^{n + 1} }}} \quad \quad ;\quad \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{2^{n + 1} }}} = \sum\limits_{n = 0}^{ + \infty } {\left( {\frac{1}{2}} \right)} ^{n + 1} = \frac{1}{2}\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{{\left( {\frac{1}{2}} \right)^{n + 1} - 1}}{{\left( {\frac{1}{2}} \right) - 1}}} \right) = 1 \cr \Leftrightarrow \frac{1}{2}s = \frac{1}{2} + 1 \cr \Leftrightarrow s = 3 \cr} $$

sabachir
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For $|r|<1$, there is the common geometric series: $$ \sum_{k=0}^\infty ar^k=\frac{a}{1-r} \overset{\times r}{\longrightarrow} \sum_{k=1}^\infty ar^k=\frac{ar}{1-r} $$ and its derivative with respect to $r$: $$ \sum_{k=0}^\infty akr^{k-1}=\frac{a}{(1-r)^2} \overset{\times r}{\longrightarrow} \sum_{k=1}^\infty akr^k=\frac{ar}{(1-r)^2} $$ Your series is a combination of these two.

robjohn
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