Proving the Maximal Member in a Converging Sequence An > 0 to 0

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An is a bounded sequence which converges to 0.
An>0 for every n.
prove that in this sequence we have a maximal member
??

again its obvious because if a sequence is descending and it converges from a positive number
to 0.
so 0 is the larger lower bound
and our first member must be the larger

how to transform it to math?
 
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If it converges to 0, then finitely many members of the sequence take on a different value (that number may be small or large but it is still finite) and the rest of the terms approach 0.
 
how to prove it mathematically?
 
By combining what you and I wrote and using proper notation
 
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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