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I'm trying to verify the solution of St.Peterburg paradox: http://en.wikipedia.org/wiki/St._Petersburg_paradox#Solutions_of_the_paradox when W = c = 2.

$$\sum_{t=1}^{\infty}2^{-t}\left(\ln\left(W+2^{t-1}-c\right)-\ln W\right)=0$$

In terms of math, I need to check if the LHS of the expression above equals to 0 when W=c=2.

$$\Leftrightarrow\sum_{t=1}^{\infty}2^{-t}\left(\left(t-1\right)\ln2-\ln2\right)=0\Leftrightarrow$$

$$\Leftrightarrow\sum_{t=1}^{\infty}2^{-t}\left(t-2\right)\ln2=0\Leftrightarrow\sum_{t=1}^{\infty}2^{-t}\left(t-2\right)=0\Leftrightarrow\sum_{t=1}^{\infty}\frac{t}{2^{t}}-2\sum_{t=1}^{\infty}\frac{1}{2^{t}}=0\Leftrightarrow$$

$$\Leftrightarrow\sum_{t=1}^{\infty}\frac{t}{2^{t}}-2\left(\frac{0.5}{1-0.5}\right)=0\Leftrightarrow\sum_{t=1}^{\infty}\frac{t}{2^{t}}=2$$

How to prove, assuming that I didn't make a mistake previously, that $\sum_{t=1}^{\infty}\frac{t}{2^{t}}=2$.

Astudent
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    Hint: Open the frequent questions list on this site, read the first question. –  Nov 01 '14 at 13:34
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    http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn –  Nov 01 '14 at 13:35

1 Answers1

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Write that $$ \sum_{n\ge 0} x^k = \frac 1{1-x} $$ on $(-1,1)$. You can write the derivative on both sides: $$ \sum_{n\ge 1} kx^{k-1} = \frac 1{(1-x)^2} \\\implies \sum_{n\ge 1} kx^{k} = \frac x{(1-x)^2} $$

Now put $x = \frac 12$ to get $$\sum_{n\ge 1} \frac k{2^k} = \frac 1{2\left(1-\frac 12\right)^2} = 2 $$

mookid
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