How to Prove a Bounded Sequence {An} Converges to L if lim inf Equals lim sup?

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Hi,

What is the proof that if series {An} is bound and its lim inf = lim sup = L, then lim A must be equal to L?
 
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What definition of "lim inf" and "lim sup" are you using?
 
Lim inf = lowest partial limit, i.e. lowest limit amongst all the limits of all the sub-series.
Lim sup = highest partial limit, i.e. highest partial limit amongst all the limits of all sub-series
 
So you are saying that if a is a limit of any subsequence, then A\le a\le A! What does that tell you?
 
a is not the limit of ANY subsequence, but the smallest of the limits of the subsequences and the largest of the limits of the subsequences.
How may I prove that the sequence itself converges to the same limit?
 
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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