Proving Series Divergence: Convergence of a_n w/o Explicit Formula

CrusaderSean
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Given a_{n} > 0 and \sum a_{n} diverges, show that \sum \frac{a_{n}}{1+a_{n}} diverges.
Since I don't have an explicit form for the series, I can't apply any of the standard tests. I'm not sure where to start on this problem. I know the criteria for convergence/divergence, namely tail end of series has to converge or cauchy criterion condition. But I don't see how that helps without knowing what series looks like. Please steer me in the right direction.
 
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If a series diverges, what happens to to its reciprocal ?
 
I would say reciprocal converges, but apparently it's not enough to bring original series to convergence... I thought about this a little more and I think I'll analyze it based on how a_{n} diverges. that is, does it go to zero, constant, or infinity as n goes to infinity and try to bound the reciprocal from below to show series diverges.
 
For a_n>1:

\frac{a_n}{1+a_n}>\frac{1}{2}

For a_n\leq 1:

\frac{a_n}{1+a_n}\geq\frac{a_n}{2}
 
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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