Question

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

Answer 1

firstly is it a g.p?

Answer 2

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 ????

Answer 3

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..

Answer 4

also .. my formula is wrong .. I think it would be somewhat like (2^(n+1) - 1)/(2^(2(2n-2))

Answer 5

\[\frac{2^{n} -1}{2^{2n}} = \frac{1}{2^{n}}-\frac{1}{2^{2n}} =(\frac{1}{2})^{n} - (\frac{1}{4})^{n}\]
\[\rightarrow 4*[\sum_{1}^{\infty}(\frac{1}{2})^{n} - \sum_{1}^{\infty}(\frac{1}{4})^{n}]\]

Answer 6

no i think your formula worked if you start at n=1

Answer 7

@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...

Answer 8

(2^5 -1)/(2^2)^4

Answer 9

got it ... thanks .. yeah my formula is just slightly of.. but thanks anyways.. :)

Answer 10

\(\large \frac 23 \) isn't ?

Answer 11

it cannot be 2/3 because it is (1 + 3/4 + ... ) and all terms are positive

Answer 12

no thats right for the summation part.

multiply by 4 and you get , sum = 8/3

Answer 13

yeah ... it wd be 8/3 ..

Answer 14

thanks both of you.. :)

Answer 15

yes it should be 8/3, I summed up dumcow's generalization, but the actualy generalization should be \( \huge \sum\limits_{k=0}^\infty \frac {2^{k+1} -1}{2^{2k}}=\frac 83\)

Answer 16

It wasn't dumcow's I didn't read the thread (fully). My apologies.