Test for convergence of the series

DarkStalker
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Homework Statement


Q)
Summation from 1 to infinity
(1+(-1)^i) / (8i+2^i)

This series apparently converges and I can't figure out why.


Homework Equations





The Attempt at a Solution



(1+(-1)^i) / i(8+2^i/i)

Taking the absolute value of the above generalization:

2/i(8+2^i/i)

Rearranging that would give:

2/i * (1/(8+2^i/i)

Now I thought that since 2/i would diverge, the entire series should diverge.

Comparing it to the geometric series (1/2^i) implies converges, but I don't know what's wrong with the above
 
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DarkStalker said:
2/i * (1/(8+2^i/i)

Now I thought that since 2/i would diverge, the entire series should diverge.
I don't understand your reasoning here. It's like saying
$$\sum_{n=1}^\infty \frac{1}{2^n}$$ diverges because you can write it as
$$\sum_{n=1}^\infty \left(\frac{1}{2}\times\frac{1}{2^{n-1}}\right)$$ and $$\sum_{n=1}^\infty \frac{1}{2}$$ diverges.
 
vela said:
I don't understand your reasoning here. It's like saying
$$\sum_{n=1}^\infty \frac{1}{2^n}$$ diverges because you can write it as
$$\sum_{n=1}^\infty \left(\frac{1}{2}\times\frac{1}{2^{n-1}}\right)$$ and $$\sum_{n=1}^\infty \frac{1}{2}$$ diverges.

Point. I just checked my textbook and it appears I'd mistakenly thought that just because Summation 1-infinity (ai+bi) equals summation 1-infinity (ai) + Summation 1-infinity (bi), I thought the same would be true for multiplication.

Thanks for clearing it up.
 
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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