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[note: also under discussion in s.o.s. math board]

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- Jan 16th 2010, 06:35 PMkingwinnerReal analysis: Limit superior proof
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[note: also under discussion in s.o.s. math board] - Jan 16th 2010, 08:08 PMDrexel28
Since we have that $\displaystyle \lim\text{ }b_n=a$ we have that for every $\displaystyle \varepsilon>0$ there exists an $\displaystyle N\in\mathbb{N}$ such that $\displaystyle N\leqslant n\implies \left|b_n-a\right|<\varepsilon\implies -\varepsilon<b_n-a<\varepsilon\implies b_n-\varepsilon<b_n<a+\varepsilon$ more pertinently we have that $\displaystyle b_N<a+\varepsilon$. But, $\displaystyle b_N=\sup\text{ }\left\{a_n:N\leqslant n\right\}$ so that $\displaystyle N\leqslant n\implies a_n\leqslant b_N<a+\varepsilon$

- Jan 16th 2010, 08:31 PMkingwinner
Sorry, I don't understand your meaning of b_n. (otherwise the proof makes sense to me)

Let (a_n) be a sequence of real numbers.

We have sup{a_n: n≥k} = sup{a_k,a_k+1,a_k+2,...} = b_k

and

lim sup{a_n: n≥k} = lim sup a_n

k->infinity

So you mean b_n = sup{a_n: n≥n} ?? - Jan 16th 2010, 08:34 PMDrexel28
- Jan 16th 2010, 09:10 PMkingwinner
OK, then I think I understand your proof. Thank you!

But now I have a general question about the way lim sup is usually defined.

Let (a_n) be a sequence of real numbers. Then we define lim sup to be

lim [sup{a_n: n≥k}] = lim sup a_n = b_k

k->∞

Here, my understanding is that the indices n and k are*independent*.

But I have seen some textbooks doing the following:

Let (a_n) be a sequence of real numbers. Then they define lim sup to be

lim [sup{a_m: m≥n}] = lim sup a_n = b_n

n->∞

i.e. they are using the same subscript "n", but "n" is the subscript in the original sequence (a_n), so they can't be independent.

Is it correct to do this?

If we use the latter definition, would this ruin our proof? (becuase at the end, I don't think we'll get the same conclusion...) - Jan 16th 2010, 09:14 PMDrexel28
That would make no sense if I understand what you're saying.

Supremum Limit -- from Wolfram MathWorld

This is the exact way you defined it, except they have n and k switched.