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My textbook says:

$$\overline{s}_n = \sup \{a_n \mid n \geq N\}$$

and $\operatorname{limsup} \{a_n\}_{n \to \infty} = \lim_{N \to \infty} \overline{s}_N$.

Also, it says: As $N$ gets larger, the sup is taken over a smaller set, so the sequence of numbers $\{\overline{s}_n\}$ is monotone decreasing.

I don't understand why this must be the case. If we have a sequence $3-1/n$, then as $n$ gets larger, don't the sequence values (and hence the supremums in each interval) get larger?

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For the sequence $3-1/n$, the supremum is $3$. Yes, as $n$ gets larger, the sequence values get larger, but the supremum stays the same. The supremum is not the thing as maximum, so supremum doesn't need to come from the set itself (indeed, $3$ does not belong to any of the elements in your sequence $3-1/n$). – Prism Apr 6 '14 at 18:20
I guess the first definition should be $\bar{s}_N=\dots$, not $\bar{s}_n=\dots$. – JiK Apr 7 '14 at 9:36
up vote 7 down vote accepted

It means "weakly" decreasing, so they may not get smaller, but they don't get larger. Let's look at $3-1/n$. \begin{align} & \sup\left\{3 - \frac1 1, 3-\frac12,3-\frac13,3-\frac14,3-\frac15,\ldots\right\} \\[8pt] & \sup\left\{ 3-\frac12,3-\frac13,3-\frac14,3-\frac15,\ldots\right\} \\[8pt] & \sup\left\{3-\frac13,3-\frac14,3-\frac15,\ldots\right\} \\[8pt] & \sup\left\{3-\frac14,3-\frac15,\ldots\right\} \\[8pt] & {}\qquad \vdots \end{align}

The sequence of suprema gets (weakly) smaller.

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I know you already accepted an answer, but nothing does it like a picture for me. Imagine our sequence does something like this where $n$ increases in the positive $x$ direction.a fluctuating sequence.

Then the limsup of this sequence would be the redline (not the axis).

The top of the peaks is red but only on one side with a horizontal line extending to the next peak.

But if the sequence looks like this,

a sequence that is eventually monotone increasing

our limsup will do this.

a red line that is horizontal and at the top of the sequence

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Very helpful, thank you! – kiwifruit Apr 7 '14 at 0:55

When I first learned about $\limsup$ and $\liminf$ I found it helpful to think of them as, respectively, "eventual least upper bound" and "eventual greatest lower bound". They are variants of the LUB and GLB that avoid taking account of any finite number of (possibly "deviant") terms of the sequence, but only record the limiting trend in how high or how low the terms get.

If you take any set $S$ of real numbers and remove some values to get a smaller set $S' \subset S$, then $\sup S' \leq \sup S$ by definition (any upper bound of $S$ is an upper bound of $S'$, in particular the LUB of $S$). The key point, which I think you are missing, is that in the definition of $\limsup a_n$, the $s_N$ are defined to be over all $n \geq N$, so for an increasing sequence like the one you gave, all the $s_N$ are equal (in this case to $3$).

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The sequence values increase but the suprema over the tails decrease because the supremum is taken over smaller and smaller sets.

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But why does this happen, if the supremum values must come from values of the sequence? Isn't the supremum the largest value of the sequence in the interval? – kiwifruit Apr 6 '14 at 17:47
@kiwifruit The supremum is not the largest value of the sequence. That would be maximum. However, maxima need not exist. In the case of the sequence $3-1/n$, the supremum is $3$ but there is no maximum because the elements in the sequence keep getting bigger and approach to $3$ in the long run. – Prism Apr 6 '14 at 18:22

This analogy may help: Imagine a piston at $+\infty$ moving leftward until it stopped by the presence of the sequence $a_1,a_2\ldots$. The place will stop is the $\sup$ of the sequence, which in your notation is $\overline s_1$, i.e., the first element of the sequence $(\overline s_n)$. Now let remove the first element $a_1$ of the original sequence, this may or may not change the $\sup$ of the sequence. If will change the $\sup$, the piston slip leftward to a new point $\overline s_2$. If not, then the piston will not move and $\overline s_2$ will be the same as $\overline s_1$. In any case $\overline s_2$ cannot be greater than $\overline s_1$. Then we remove the second element $a_2$, causing the piston to slip a little more. If we keep doing this the piston will keep slipping, but there will be a point where cannot go any further and this is what we call the limit superior of the sequence.

Now more formally the sequence $(\overline s_n)$ is non-increasing, i.e., is monotone, so the limit of the sequence always exists in the extended real numbers, and this limit, which is the infimum, is what we call the limit superior.

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