Is the Infimum of Set A Equal to Zero?

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Homework Statement


Suppose A={\frac{1}{1},\frac{1}{2},...}={\frac{1}{n}|n\in{Z^+}}

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The Attempt at a Solution


Could you take the limit of \frac{1}{n} as \infty to prove this, or would I go about it a different route?
 
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Just use the definition of an infimum. There's no need to invoke limits.
 
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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