Originally Posted by

**paupsers** Here's as far as I can get...

I know that $\displaystyle S_n=\{a_k|k \geq n\}=\{a_n, a_{n+1}, a_{n+2}...\}$

Then, how do I define $\displaystyle \sup S_n=?$ since $\displaystyle S_n$ has an infinite number of terms?

So?? The definition of supremum is interesting ONLY if there's an infinite number of elements, otherwise we can always choose the largest one by observation which then would be the maximum

Moreover, I don't understand the convention of writing $\displaystyle \limsup a_n=\inf\{\sup S_n\}$ because isn't $\displaystyle \sup S_n$ a single value, so the $\displaystyle \inf\{\sup S_n\}$ is the same value?

Yes, $\displaystyle \sup S_n$ is a single value FOR EVERY SINGLE n, and you take the infimum over all the possible n's!

The definition my book gives for $\displaystyle \limsup$ is: if $\displaystyle T$ is the set of all subsequential limits of $\displaystyle \{a_n\}$, then $\displaystyle \sup T$ is $\displaystyle \limsup \{a_n\}$. I don't understand how this definition is equivalent to $\displaystyle \limsup a_n=\inf\{\sup S_n\}$.

If t is a partial limit then there's an infinite subsequent which converges to t and, thus there's an infinite number of elements of the original sequence which converge to t.

don't you try some examples? Take for example $\displaystyle \{a_n\}=\left\{-1,1,1,-1,1,\frac{1}{2},-1,1,\frac{1}{3},...., -1,1,\frac{1}{n},....\right\}$

In this sequence, we'll get for infinite n's that $\displaystyle \sup S_n = 1\Longrightarrow \inf \sup S_n=1\,,\,\,and\,\,\inf S_n=-1\Longrightarrow \sup \inf S_n=-1$, so the lim sup is 1 and the lim inf is -1.

Tonio

Does anyone understand what I'm asking? Lol. I'm just looking for a way to show that the two definitions are equivalent.