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Given a series:

$\frac{1}{3*2} +\frac{4}{3*2^2} + .... + \frac{3n-2}{3*2^n}$

Find the sum of this series?

My attempt:

I think I should try to find the first 5 terms to find the pattern, such that:

$S_1= \frac{1}{6}$

$S_2= \frac{1}{2}$

$S_3= \frac{19}{24}$

$S_4= 1$

$S_5= \frac{109}{96}$

But it seems that didn't help me to find its pattern. Can someone provide a better way to find the solution?

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  • $\begingroup$ $S_3=\frac{7}{24}$, $S_4=\frac{10}{48}$ $\endgroup$ – sirous Nov 25 '17 at 15:33
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$$\sum_{n=1}^\infty \frac{3n-2}{3\cdot 2^n} = \sum_{n=1}^\infty \frac{n}{2^n} - \frac{1}{3}\sum_{n=1}^\infty \frac{1}{2^{n-1}} = -\frac{2}{3} + \sum_{n=1}^\infty \frac{n}{2^n}$$ since both sums converge. To evaluate the latter sum, note that $$\frac{1}{1-x} = \sum_{i=0}^\infty x^i,$$ so that $$\left(\frac{1}{1-x}\right)' = \frac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1} = 1 + 2x + 3x^2 + \cdots.$$ Then set $x=\frac{1}{2}$, and $$2 = \frac{1/2}{(1-1/2)^2} = \frac{1}{2}\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n-1} = \sum_{n=1}^\infty \frac{n}{2^n}.$$ Finally, $$\sum_{n=1}^\infty \frac{3n-2}{3\cdot 2^n} = -\frac{2}{3} + \sum_{n=1}^\infty \frac{n}{2^n} = -\frac{2}{3}+2 = \frac{4}{3}$$ and you are done.

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  • $\begingroup$ Then the last word from your solution is equal with 2? $\endgroup$ – akusaja Nov 25 '17 at 15:50
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i think such a formula is difficult to find it is $$\frac{1}{3}\left(\frac{2^{2+n}-3n-4}{2^n}\right)$$

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Try to sum the partial summation

$ \sum_{n=1}^m (3n-2)/(3* 2^n)=\sum_{n=1}^m n2^{-n} -\frac 2 3 \sum_{n=1}^m 2^{-n}=\left. \left(\sum_{n=1}^m n x^{n} -\frac 2 3 \sum_{n=1}^m x^{n}\right)\right|_{x=1/2}$ So I think that it better than my previous thought. You can take the limit already and then:

$ \left. \left(\sum_{n=1}^\infty n x^{n} -\frac 2 3 \sum_{n=1}^\infty x^{n}\right)\right|_{x=1/2}=\left.\left(x \frac{d}{dx}\left(\frac{x}{1-x}\right)-\frac 2 3\frac{x}{1-x}\right)\right|_{x=1/2} =\left.\left(\frac{x}{(1-x)^2}-\frac 2 3\frac{x}{1-x}\right)\right|_{x=1/2} =4/3 $

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  • $\begingroup$ Can you show me how to get the solution? Thanks $\endgroup$ – akusaja Nov 25 '17 at 15:33
  • $\begingroup$ I have corrected my previous thought, but then I also have found out that I was previous considering the series from 0 e not from 1 $\endgroup$ – Alessandro Mininno Nov 25 '17 at 16:16
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Suppose $S = \frac{1}{3*2} +\frac{4}{3*2^2} + .... + \frac{3n-2}{3*2^n}$ then $3S = \frac{1}{2} +\frac{4}{2^2} + .... + \frac{3n-2}{2^n}$ and now $$\frac{3S}{2} = \sum_{n=1}^{\infty}\frac{3n-2}{2^{n+1}}$$Now Subtract $3S- \frac{3S}{2} = \frac{1}{2} + \sum_{n=1}^ {\infty} \frac{3}{2^{n+1}}$ and you get the answer.

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