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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.

Martin Sleziak
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Iching Sebastian Quares
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    differentiate both side of $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$, then $z=1/2$ – reuns May 17 '16 at 03:34
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    Assume to toss a fair coin until a head shows up. How many tosses do you need, on average? – Jack D'Aurizio May 17 '16 at 03:54
  • ($2^{-n-1}$ 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) – reuns May 17 '16 at 04:10
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    You can find several posts about this series on this site: http://math.stackexchange.com/questions/337937/why-sum-k-1-infty-frack2k-2 http://math.stackexchange.com/questions/441481/why-does-sum-n-0-infty-fracn2n-converge-to-2 http://math.stackexchange.com/questions/674220/a-simple-series http://math.stackexchange.com/questions/1330493/how-do-you-prove-sum-frac-n2n-2 http://math.stackexchange.com/questions/757263/how-to-find-answer-to-the-sum-of-series-sum-n-1-infty-fracn2n – Martin Sleziak Oct 06 '16 at 02:45

2 Answers2

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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 $S=2$

Bijesh K.S
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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 $|x|<1$.

kmitov
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