Infinite Series Homework: Determine Convergence/Divergence

mreaume
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Homework Statement



Determine whether the series diverges or converges.

(1+2) / (1+3)+ ((1+2+4)/(1+3+9))+ ((1+2+4+8)/(1+3+9+27)) + ...

The Attempt at a Solution



I have split up the series into two (denominator and numerator):

an = (1+2) + (1+2+4) + (1+2+4+8)+... = (1)n + 2n + 4(n-1) + ...
bn = (1+3) + (1+3+9)+... = (1)n + (3)n + (9)(n-1)+... = (1)n + 3n + 9(n-1) + ...I don't know how to keep going. I suspect that the ratio test will come in handy later but am not sure how to apply it with the given series above. Any help would be appreciated. Thanks.
 
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How is evaluating an and bn supposed to help? an/bn is not the same as the partial sum of the series you have been given.

You might want to think about a short way to write down
1+2+4+8+16+...
for any finite term.
 
Well we can write 1+2+4+8+... as

the sum from n=0 to n=inf of 2^n.

And similarly we can write 1+3+9+...

as the sum from n=0 to n=inf of 3^n.

So can we say that the series is (2/3)^n? From n=0 to n=inf?
 
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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