# Real analysis: Limit superior proof

• Jan 16th 2010, 06:35 PM
kingwinner
Real analysis: Limit superior proof

[note: also under discussion in s.o.s. math board]
• Jan 16th 2010, 08:08 PM
Drexel28
Quote:

Originally Posted by kingwinner

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

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 PM
kingwinner
Quote:

Originally Posted by Drexel28
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 $\displaystyle N\leqslant n\implies a_n\leqslant b_N<a+\varepsilon$

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 PM
Drexel28
Quote:

Originally Posted by kingwinner
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

So b_n = sup{a_n: n≥n} ??

I'm sorry. I guess I'm just so used to using n as a subscript. Change it to k.
• Jan 16th 2010, 09:10 PM
kingwinner
Quote:

Originally Posted by Drexel28
I'm sorry. I guess I'm just so used to using n as a subscript. Change it to k.

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 PM
Drexel28
Quote:

Originally Posted by kingwinner
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...)

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.