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Help needed... sum of series = 1 + (3/4) + (7/16) + (15/64) + (31/256) + .. infinity

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firstly is it a g.p?
okk I got this much that = 1 + (2^2 -1)/2^2 + (2^3 -1)/(2^4) + (2^4 - 1)/2^16 + ... = \[2^2 * \sum_{0}^{\infty} (2^n-1)/(2^2)^n\] Now what ????
its not a gp.. its a combination .. think someone with little practice can answer this .. I have lost touch and hence have forgot the method to sum these..

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also .. my formula is wrong .. I think it would be somewhat like (2^(n+1) - 1)/(2^(2(2n-2))
\[\frac{2^{n} -1}{2^{2n}} = \frac{1}{2^{n}}-\frac{1}{2^{2n}} =(\frac{1}{2})^{n} - (\frac{1}{4})^{n}\] \[\rightarrow 4*[\sum_{1}^{\infty}(\frac{1}{2})^{n} - \sum_{1}^{\infty}(\frac{1}{4})^{n}]\]
no i think your formula worked if you start at n=1
@dumbcow .. no I mistyped the formula .. look at the series ... you won't get 7/16 .. or 31/256 ... it is a little more complicated...
(2^5 -1)/(2^2)^4
got it ... thanks .. yeah my formula is just slightly of.. but thanks anyways.. :)
\(\large \frac 23 \) isn't ?
it cannot be 2/3 because it is (1 + 3/4 + ... ) and all terms are positive
no thats right for the summation part. multiply by 4 and you get , sum = 8/3
yeah ... it wd be 8/3 ..
thanks both of you.. :)
yes it should be 8/3, I summed up dumcow's generalization, but the actualy generalization should be \( \huge \sum\limits_{k=0}^\infty \frac {2^{k+1} -1}{2^{2k}}=\frac 83\)
It wasn't dumcow's I didn't read the thread (fully). My apologies.

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