Is s_m the Same as s_n in Advanced Calculus Problem?

[mynameisfunk]

Homework Statement

See attachment

I am confused by what they mean on part (a) are these sequences $$s_m$$ and $$s_n$$ the same sequence and this is what i should be showing?

 

[Inferior89]

Edit: Misread the question.

 

[mynameisfunk]

wouldnt that make (a) false then? but a goes to infinity... truncated on infinity?

 

[Inferior89]

Misread the question. Thought it said m < n... Sorry.

 

[mynameisfunk]

also, if someone could show me show to write part a in LateX i would appreciate it

 

[Inferior89]

I don't know the answer to the question but I can show you the latex.

$$\mathop {\lim \sup }\limits_{n \to \infty}} s_n = \mathop {\lim}\limits_{n \to \infty}} \left( \sup \lbrace s_m | m \geq n \rbrace \right)$$

 \mathop {\lim \sup }\limits_{n \to \infty}} s_n = \mathop {\lim}\limits_{n \to \infty}} \left( \sup \lbrace s_m | m \geq n \rbrace  \right)

 

[mynameisfunk]

with the inequality $$m \leq n$$

 

[Inferior89]

$$\mathop {\lim }\limits_{n \to \infty}} \mathop {\sup }\limits_{m \geq n}} s_m$$

\mathop {\lim  }\limits_{n \to \infty}} \mathop {\sup }\limits_{m \geq n}} s_m

 

[HallsofIvy]

Yes, $s_n$ and $s_m$ refer to the same sequence, just numbered differently.

As for $\lim_{n\to\infty} sup s_n= \lim_{n\to\infty}\{sup s_m|m\ge n\}$
look at some simple examples. if n= 1, then $\{s_m|m\ge 1\}$ is just the sequence itself. If n= 2, then $\{s_m|m\ge 2\}$ is all terms of the sequence except the first. In general, $\{s_m|m\ge n\}$ is the set of all terms of the sequence except those before $s_n$. "$\{sup s_n|m\ge 1\}$ is the supremum (greatest lower bound) of all terms in the sequence beyond a certain point.

In a special case, suppose $\{a_n\}$ converges to A. Then, given $\epsilon> 0$ there exist N such that if n> N, $|a_n- A|< \epsilon$ so there are no members of the sequence, beyond n= N, that are larger than $A+ \epsilon$ and it is easy to see that, as n goes to infinity, the supremums of $\{s_n|m\ge n\}$ must go to A.

Suppose the sequence has a subsequence that converges to A and a subsequence that converges to B. Then no matter how large n is, there exist numbers in the sequence beyond n that are close to A and numbers close to B. The supremum will be close to whichever of A or B is larger. In the limit, the supremum will be the larger of A and B.

In general lim sup is the supremum of the set of all sub-sequential limits of the sequence. That is, you determine all numbers to which sub-sequences converge and find their supremum.

 

[mynameisfunk]

Thanks