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&gt; 0 there exist N such that if n> N, |a_n- A|&lt; \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