Disprove a convergence question

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i know that An->1

i need to prove that (An)^n ->1

but when i construct limit
lim (An)^n
n->+infinity

the base goes to 1 and the power goes to + infinity

that is not solvable

i get 1^(+infinity) which says that there is no limit
what do i do in this case in order to disprove that (An)^n->1

??
 
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All you need is a counterexample to show that
\lim_{n \rightarrow \infty} (a_n)^n = 1 isn't true.

You need a sequence {a_n} whose limit is 1 but for which the limit above isn't 1.
 
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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