I know that the series converges by d'Alembert ratio test, where $\lim\left ( \frac{A_{n+1}}{A_{n}} \right )= \frac{1}{2}$, but I don't know how to calculate the sum of the serie. Thanks for the help.
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2$\begingroup$ differentiate both side of $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$, then $z=1/2$ $\endgroup$– reunsMay 17, 2016 at 3:34
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3$\begingroup$ Assume to toss a fair coin until a head shows up. How many tosses do you need, on average? $\endgroup$– Jack D'AurizioMay 17, 2016 at 3:54
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$\begingroup$ ($2^{-n-1}$ is the probability to have $n$ faces followed by a head, and $\mathbb{E}(Y) = \sum_n n 2^{-n-1}$ where $Y$ is the number of tosses before a head) $\endgroup$– reunsMay 17, 2016 at 4:10
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4$\begingroup$ You can find several posts about this series on this site: math.stackexchange.com/questions/337937/… math.stackexchange.com/questions/441481/… math.stackexchange.com/questions/674220/a-simple-series math.stackexchange.com/questions/1330493/… math.stackexchange.com/questions/757263/… $\endgroup$– Martin SleziakOct 6, 2016 at 2:45
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2 Answers
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\begin{align*} S&= \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+.....\\ \frac{1}{2}S&=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\\ S-\frac{1}{2}S&=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......\\ S-\frac{1}{2}S&=1 \end{align*} Thus $S=2$
answered May 17, 2016 at 3:40
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$\sum_{k=0}^\infty x^n = \frac{1}{1-x}$ for $|x|<1$. Then $\sum_{k=1}^\infty nx^n = x\times \left(\frac{1}{1-x}\right)'$ for $|x|<1$.
