Trouble understanding difference between the following two results involving the number of terms in the set $\{n: s_n \gt \lim \sup s_n\}$

Question

There are two results written in Ross' book on Elementary Analysis, which state the following -

If $L = \limsup s_n \neq \infty$, then for every $\alpha > L$, the set $\{n: s_n \gt \alpha\}$ is finite.

The set $\{n : s_n \gt \limsup s_n\}$ can be infinite. For example, consider ($s_n$) where $s_n = \frac{1}{n}$.

In the second example, $\limsup s_n = \lim s_n = 0 \neq \infty$. Thus, by result 1, the set $\{n : s_n \gt \limsup s_n\}$ should be finite. The only difference between the two results seems to be the existence of all $\alpha$ such that $\alpha \gt \limsup s_n$(in the first result). But why would that make such a significant difference? I am unable to understand this intuitively. Thank you for your time.

Answer 1

In the first result, $\alpha$ is considered as being a value that is greater that the limit superior of the given sequence. In the example given, consider the graph of the real function $y=f(x)=\frac{1}{x}$. As you can clearly see, the graph is asymptotic to the x-axis. Here as you have pointed out yourself , the limit superior is 0, which is analogous to the x-axis in this case. The second result says that the number of terms above the x-axis are infinite, and this is actually due to the fact the function is asymptotic to the x-axis. However, what the first result says is, if you take a new line, parallel to the axis axis, but just lying a teensy-bit above it, then the number of terms above the line will no longer be be infinite, as while $y=0$ is an asymptote, $y=c$ for $c>0$ is no longer one. Instead, it intersects the function at a particular point. And as the function is decreasing, the number of terms above $y=c$ are finite, only those terms that appear before the intersection point.

To think intuitively, an asymptote no longer remains an asymptote, if you shift it, parallel to itself, infinitesimally towards the curve it is an asymptote of.

I know that many people would point out that I am completely mixing up a real function with a sequence, but in my opinion this type of analogy helps in the intuitive process.

Answer 2

Result 1 does not state the set $\{n : s_n \gt \limsup s_n\}$ is finite, it states that for every $\alpha > L=\limsup s_n$, the set $\{n: s_n \gt \alpha\}$ is finite. These two claims are completely different.

In result 1, $\alpha$ has to be greater than $\limsup s_n=0$, pick any $\alpha>0$, there exists a finite $N$ such that for all $n\ge N,\frac1n<\alpha$, so the set $S_\alpha=\{n: s_n=\frac1n \gt \alpha\}$ must be finite. You cannot pick $\alpha=\limsup s_n=0$.

In result 2, the set $S=\{n: s_n=\frac1n> \limsup s_n=0\}$ is indeed infinite. In fact, $S=\lim_{\alpha\to0}S_\alpha$. But just because $S_\alpha$ is finite for every $\alpha>0$ does not mean $S=\lim_{\alpha\to0}S_\alpha$ is also finite.