# Arithmetics of superior limits

• Dec 13th 2013, 08:48 AM
davidciprut
Arithmetics of superior limits
• Dec 13th 2013, 10:13 AM
Plato
Re: Arithmetics of superior limits
It is really hard to give you an answer. There two reasons: first it appears from your #2 that you do not even understand how limit superiors are defined; second the notations vary widely from text to text.

That said, let $(s_n)\subset\mathbb{R}$ be a sequence then:
$\lim \,\sup \left( {{s_n}} \right) = {\lim _{N \to \infty }}\sup\left\{ {{s_n}:n > N} \right\}$

I will give some examples. The first two are bounded sequences.

Let $s_n=1-n^{-1}$ then $\lim \,\sup \left( {{s_n}} \right)=1$

Let $t_n=n^{-1}$ then $\lim \,\sup \left( {{t_n}} \right)=0$. This example maybe where the idea of the tail comes from.

Let $u_n=\begin{cases}n^{-1} &: n\text{ is odd}\\n &: n\text{ is even}\end{cases}$ then $\lim \,\sup \left( {{u_n}} \right)=\infty$

For #1, what do you know about those sequences? Are they bounded?
• Dec 13th 2013, 10:36 AM
davidciprut
Re: Arithmetics of superior limits
So, if I understood it correctly, it's the same idea with the partial limits, in other words we have a sequence, and we have the set S, the set of all partial limits of the sequence and the greatest is equal to the limit of supun, and that is actually the same thing with the expression I wrote, right ? For example the third example that you gave has two subsequences , one converges to infinity, and the other to 0, so S={0, infinity} and infinity is superior therefore we say limsup(un)=infinity (I don't know how to write the sign)

For the first question, yes , they are bounded.
• Dec 13th 2013, 11:42 AM
Plato
Re: Arithmetics of superior limits
Quote:

Originally Posted by davidciprut
For the first question, yes , they are bounded.

If each sequence is bounded, then $(a_n+b_n)$ is bounded and the $\lim\sup$ exists.

Look back at reply #2. Let $\tau_N=\sup\{t_n: n>N\}.$
Is it clear that the sequence $(\tau_n)$ is non-increasing and bounded below by $\lim\sup (t_n)~?$

Also let $\sigma_N=\sup\{s_n: n>N\}.$.
Is it also clear that the sequence $(\sigma_n)$ is non-increasing and bounded below by $\lim\sup (s_n)~?$

Now let $\alpha=\lim\sup (a_n),~\beta=\lim\sup (b_n)~\&~\gamma=\lim\sup (a_n+b_n)$.
You want to prove that $\gamma\le\alpha+\beta$.
By contradiction, what happens if $\alpha+\beta< \gamma~?$
• Dec 13th 2013, 12:05 PM
davidciprut
Re: Arithmetics of superior limits
Shouldn't that be alpha + beta is greater or equal to gamma?
• Dec 13th 2013, 12:13 PM
Plato
Re: Arithmetics of superior limits
Quote:

Originally Posted by davidciprut
Shouldn't that be alpha + beta is greater or equal to gamma?

YES, and I made the edit.
• Dec 13th 2013, 12:38 PM
romsek
Re: Arithmetics of superior limits
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