Calculate the sum of the series

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so i am having trouble calculating the sum of the series of
(1+2^x)/(3^x) from 1 to infinity
i rearranged it and made a geometric series where r =2/3 and got that it converges toward 2. however the answer is 2.5? This problem shouldn't need any tests as its it a section of the book that hasant taught any of the tests, except for geometric so what am I doing wrong. I have been stuck on this for hours now? :(
 
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Let's see details for how you went about getting your result.
 
shemer77 said:
so i am having trouble calculating the sum of the series of
(1+2^x)/(3^x) from 1 to infinity
i rearranged it and made a geometric series where r =2/3 and got that it converges toward 2. however the answer is 2.5? This problem shouldn't need any tests as its it a section of the book that hasant taught any of the tests, except for geometric so what am I doing wrong. I have been stuck on this for hours now? :(

You forgot about the 1 in the numerator:

\sum \frac{1 + 2^x}{3^x} = \sum \left(\frac{1}{3^x} + \frac{2^x}{3^x} \right) = \sum \frac{1}{3^x} + \sum \left(\frac{2}{3}\right)^x = ?
 
Borhok has the right method again.
Shemer77, would you let us know if you've gotten the right answer from Bohrok's information or still have further questions? Thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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