Proving that ##lim sup s_n = lim s_n##

[Hall]

Homework Statement

Main body

Relevant Equations

Main body

Let $v_N = \sup \{ s_n : n \gt N \}$. If $lim \sup s_n = \lim v_N = L$, then for $\epsilon /gt 0$, we have $N$ such that
$$
m \gt N \implies v_m \lt L + \varepsilon$$
$$
n\gt m \implies s_n \lt L + \epsilon$$
Therefore, $\lim s_n = L$.

I don't very much understand limit superior, but simply writing it as $v_N$ allowed me to carry out the usual procedure.

 

[jambaugh]

To understand lim sup, take a straightforward example, say:
a_n = (-1)^n\frac{n}{n-1}
This alternating sequence does not converge to zero, so it has no limit. In the limit, it bounces between "close to" 1 and "close to" -1. Work out $A_n = \mathop{sup}_{k\ge n} a_k$. That will be $a_n$ for even n and $a_{n+1}$ for odd n. As the magnitude is decreasing the first positive value in the tail will be the maximum.

Then note that the limit of $A_n$, which is the lim sup of $a_n$ is 1.
(and the lim inf would then be -1).

That should help clarify the definition.

 

[FactChecker]

What if the limit does not exist? Does that mean that the lim sup does not exist?

 

[Hall]

To understand lim sup, take a straightforward example, say:
a_n = (-1)^n\frac{n}{n-1}
This alternating sequence does not converge to zero, so it has no limit. In the limit, it bounces between "close to" 1 and "close to" -1. Work out $A_n = \mathop{sup}_{k\ge n} a_k$. That will be $a_n$ for even n and $a_{n+1}$ for odd n. As the magnitude is decreasing the first positive value in the tail will be the maximum.

Then note that the limit of $A_n$, which is the lim sup of $a_n$ is 1.
(and the lim inf would then be -1).

That should help clarify the definition.

Yes, it has clarified that if the sequence is decreasing then $\sup \{s_n : n \in \mathbf{N} \} \geq \lim_{N \to \infty} \sup \{s_n : n \gt N \}$

 

[Hall]

What if the limit does not exist? Does that mean that the lim sup does not exist?

I have only showed one way implication, that if limit superior exists then limit of original sequence is equal to it.

If the limit doesn't exist, then also limit superior might exist as in the case of $(-1)^n$ (but here I'm simply assuming that limit superior is, nothing but, limit of a subsequence)

 

[FactChecker]

I have only showed one way implication, that if limit superior exists then limit of original sequence is equal to it.

If the limit doesn't exist, then also limit superior might exist as in the case of $(-1)^n$ (but here I'm simply assuming that limit superior is, nothing but, limit of a subsequence)

The statement of what you proved must also include the assumption that the limit does exist.

 

[Hall]

The statement of what you proved must also include the assumption that the limit does exist.

Okay.

So, can I take limit superior totally as a limit of a subsequence?

 

[FactChecker]

Okay.

So, can I take limit superior totally as a limit of a subsequence?

if the limit exists, then lim sup = lim

 

[Hall]

Rudin has made a few things clear.

Given a sequence $(s_n)$. If it’s subsequence $(s_{n_k} )$ converges to $x$, then we define $E$ to be a set of all those $x$ s, that is all the limits of all possible subsequences. We write
$$
\lim \sup s_n = \sup E
$$
$$
\lim \inf s_n = \inf E$$.

Prove that if a sequence $(s_n)$ has a limit, then $\lim \inf s_n = \lim s_n = \lim \sup s_n$.

Proof: Since, $\lim (s_n)= s$, we can prove that all the subsequences of $(s_n)$ will converge to $s$. So, the set E contains just one element: $s$. So, $\sup E = \inf E = s$.

So,
$$\lim \inf s_n = s = \lim \sup s_n$$.

That was to be demonstrated.