Is the Sequence {a_n} Convergent?

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Homework Statement


Let {a_n} be a sequence | (a_n+1)^2 < (a_n)^2, 0 < (a_n+1) + (a_n). Show that the sequence is convergent


Homework Equations



n/a

The Attempt at a Solution



So I am feeling like monotone convergence theorem is the way to go there. It seems to me that (a_n+1)^2 < (a_n)^2 would imply the sequence is decreasing, but I do not know what to do with 0 < (a_n+1) + (a_n) to show it is bounded.
 
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Without the second inequality, you could construct series like 1+1/2, -1-1/3, 1+1/4, -1-1/5, ... - it has to be bounded based on the first inequality alone, but this is not sufficient for convergence.
With both inequalities, you can rule out sign switches of a_n and get monotony.
 
I do not understand how the first inequality shows that a_n is bounded.
 
##a_{n+1}^2 < a_n^2## is equivalent to ##|a_{n+1}| < |a_n|##, which leads to ##|a_{n}| < |a_0|\, \forall n \in \mathbb{N}##.
 
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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