Upper and lower limit proof (liminf/sup)

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Jesssa
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Hey,
So I have been working on this question for quite a while now and I'm at this point.
I just wanted to check if everything was okay, I never feel confident with questions like this. Here is the question and my working,

http://img404.imageshack.us/img404/8436/afafafa.jpg

Is there anything wrong?

Thanks in advanced.
 
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Are there are conditions on Xn and Yn? The inequality is not true in general. If we set Xn - Yn = a = const, then Xn + Yn = 2, so lim inf {Xn + Yn} = lim sup {Xn + Yn} = a, but lim inf Xn and lim sup Xy may not exist.
 
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The conditions are that they are both bounded real sequences,

Hmm
 
OK, bounded makes it correct.

You have the first part correctly.

The second, however, is wrong. You show that Ix + Sy <= Sx + Sy. That's true, but that does not mean that Ix + Sx <= Sw.
 
I have found another way, do you think this works?

http://img211.imageshack.us/img211/1769/89852645.jpg

the last part of the first line sup(inf(x)+y) = inf(x) + sup(y), because inf(x) will just be some number,
 
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This looks good to me.
 
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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