Upper and Lower Sequence of a Convergent Sequence

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if my sequence is

1/2,2/3,3/4,4/5 ..

why the upper sequence is 1

and lower sequence is 1 too ??

1 is not ever a member of a sequence.

i got the idea that upper sequence is a sub sequence constructed from the highest members
the closest to the upper bound

1 is not even in the sequence??
 
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First, you don't mean "upper sequence"- there is no such thing. You may mean "least upper bound" of this sequence. Yes, every number in the sequence is less 1 so 1 is an upper bound on the sequence. Further, the sequence clearly converges to 1 (even though 1 itself is not in the sequence) so no number less than 1 can be an upper bound. 1 is the least upper bound. 1/2 is the smallest number in the sequence and so is the "greatest lower bound".

You have posted other questions about "lim sup" and "lim inf"- in case that is related to this" because this sequence converges to 1, the "set of subsequence limits" contains only the number 1. lim inf and lim sup for this sequence are both 1.
 
there is such a thing as lower /upper sequence

http://img383.imageshack.us/img383/6550/19840671gt7.gif

i got it from this article
http://pages.pomona.edu/~gk014747/teaching/Fall2008/math101_Fall2008_L16.pdf
 
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Ah, thank you for posting that. What the website you reference says is that you form the "upper sequence" by taking the least upper bound of all numbers in the sequence at or past[/starting an. In this case, you have an increasing sequence that converges to 1. The least upper bound of the entire set of numbers, {1/2, 2/3, 3/4, 4/5, ...} is 1. The least upper bound of "all except the first number", {2/3, 3/4, 4/5, ...} is also 1. Remember that the "least upper bound" of a set does not have to be in the set. Strictly speaking the "upper sequence" is not the number "1" but rather the constant sequence {1, 1, 1, 1, ...}.

The "lower sequence" is the greatest lower bound of all number at or past an. In this case, the greatest lower bound of {1/2, 2/3, 3/4, ...} is its minimum, 1/2. The greatest lower bound of {2/3, 3/4, ...} is its minimum, 2/3. In other words, the "lower sequence" is just {1/2, 2/3, 3/4, ...} itself.
 
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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