Question reagarding liminf definition

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regarding this definition

<br /> <br /> a = \lim \bigg( \inf \{ a_k | k\geq n\} \bigg) <br />

the sequence
<br /> <br /> \inf \{ a_k | k\geq n \}<br />

is non decreasing . its inf gets bigger or not changing in each following sequence
so its limit is its least upper bound

am i correct??
 
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Yes.
 
what is the relations between this liminf and
all the members in the sequence

is it bigger or smaller then all of them?
 
There is no requirement that a "liminf" be smaller or larger than all members of the sequence. The liminf of a sequence is the least upper bound of all subsequential limits.

For example, suppose a_n is (n-6)/2n if n is odd, -(n-6)/2n if n is even. Then {a_n} is {-5/2, 1, -1/2, 1/4, -1/10, 0, 1/14, -1/8, ...}. For n odd, we have a sequence that converges to 1/2. For n even, we have a sequence that converges to -1/2. The liminf is the smaller of those, -1/2 but there is a number in the sequence less than -1/2. The limsup is 1/2 but there is a term of the sequence larger than 1/2. We can change any finite number of terms in a sequence with changing any subsequential limits so there cannot be any relation between the limit or liminf or limsup and individual terms of the sequence.
 
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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