# Thread: Real analysis: Limit superior proof

1. ## Real analysis: Limit superior proof

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

2. Originally Posted by kingwinner

[note: also under discussion in s.o.s. math board]
Since we have that $\lim\text{ }b_n=a$ we have that for every $\varepsilon>0$ there exists an $N\in\mathbb{N}$ such that $N\leqslant n\implies \left|b_n-a\right|<\varepsilon\implies -\varepsilon more pertinently we have that $b_N. But, $b_N=\sup\text{ }\left\{a_n:N\leqslant n\right\}$ so that $N\leqslant n\implies a_n\leqslant b_N

3. Originally Posted by Drexel28
Since we have that $\lim\text{ }b_n=a$ we have that for every $\varepsilon>0$ there exists an $N\in\mathbb{N}$ such that $N\leqslant n\implies \left|b_n-a\right|<\varepsilon\implies -\varepsilon more pertinently we have that $b_N. But, $b_N=\sup\text{ }\left\{a_n:N\leqslant n\right\}$ so that $N\leqslant n\implies a_n\leqslant b_N
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} ??

4. 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.

5. 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...)

6. 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.