I am seriously having trouble understanding the meaning of upper and lower limits. Can someone give me easy-to-follow examples and explanations for the following?
Def: Let $\{s_n\}$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ such that $s_{n_k} \to x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits plus possibly the numbers $+\infty$ and $-\infty$.
Putting $s^* =\sup(E)$ and $s_* = \inf(E)$, we call them the upper and the lower limits. We also use the notation $$\lim_{n \to \infty} \sup (s_n) = s^*$$ $$\lim_{n \to \infty} \inf (s_n) = s_*$$
This is my understanding.
a), $\{s_n\}$ is a sequence, and its subsequence (there are $\infty$ many patterns) $\{s_{n_k}\}$ has many different limits. That's why it's possible to have a set $E$ which may contain more than one limit.
b), The limit of $\{s_n\}$ is not necessarily the limit of all subsequence of $\{s_n\}$.
c), The "largest" number $x$ in $E$ is the "upper limit." The analogue of the lower limit would be the "smallest".
Now this is what I'm getting confused.
What does $\sup (s_n)$ mean? Why do we have to take the limit as $n \to \infty$ to get the supremum? Isn't $\sup (s_n)$ already the supremum of $E$?
Another thing, can't there be a subsequence of $\{s_n\}$ that has a limit point greater than that of $\{s_n\}$?
I may be asking some weird questions, but that's just because this idea still doesn't click with me.
The reason I am asking this question in the first place is, because I thought I understood it but I couldn't uderstand the following.
Consider the series $$ \frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3} + \cdots $$
I was working on the ratio test and I miserably failed to understand the text.
My claim is that $$a_n = \frac{3^n+2^n}{6^n}$$ so the ratio test would give me $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{6} \left({\frac{3+2(\frac{2}{3})^n}{1+(\frac{2}{3})^n}}\right)$$
But the book tells me that $$\lim_{n \to \infty} \inf {\frac{a_{n+1}}{a_n}} = \lim_{n \to \infty} (\frac{2}{3})^n$$ $$\lim_{n \to \infty} \sup{\frac{a_{n+1}}{a_n}} = \lim_{n \to \infty} (\frac{3}{2})^n$$ $$\lim_{n \to \infty} \inf{\sqrt[n]{a_n}} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{3^n}}$$ $$\lim_{n \to \infty} \sup {\sqrt[n]{a_n}} = \lim_{n \to \infty} \sqrt[2n]{\frac{1}{2^n}}$$
I've spent 4 hours trying to understand this but I have absolutely no idea how and why these numbers popped up. Especially the "2n th" roots for the root test.
I am asking a lot, but it's really driving me mad and I need help 😛
Best Answer
In this context, I believe that they mean $\sup_{k\geq n}(s_k)$ so that the expression for the upper limit is $\lim_{n\rightarrow \infty} \sup_{k\geq n}(s_k)$. Note that you can have some numbers in your sequence larger than the upper limit. In fact, every number in the sequence can be larger than the upper limit, take the sequence $\{\frac{1}{n}\}$. Its upper limit is 0, but every number in the sequence is larger than 0. Another important property is that a sequence converges in $\mathbb{R}$ if and only if the upper and lower limits are in $\mathbb{R}$ and coincide.
Edit: The supremum of the set $\{\frac{1}{n}\}$ is 1, but the upper limit is not 1. To compute the upper limit, the idea is this. Take your sequence $\{s_n\}$ and form a new sequence, call it $\{t_n\}$ (they all may be infinite). Define $t_1$ to be the sup of $\{s_1,s_2,s_3\ldots\}$. Define $t_2$ to be the sup of $\{s_2,s_3,s_4\ldots\}$. $t_3$ to be the sup of $\{s_3,s_4,s_5\ldots\}$. and so on. Now check that $\{t_n\}$ is a decreasing sequence (you are throwing away potentially larger terms). Therefore it has a limit (maybe infinite). This limit is the upper limit.
Edit 2: Now for the lower limit. Start with the given sequence $\{s_n\}$. The lower limit is $\lim_{n\rightarrow \infty}\inf_{k\geq n}(s_k)$. Let's unravel this definition as we did for the upper limit. Define a new sequence of numbers $\{t_n\}$ (possibly including $-\infty$) as follows. Define $t_1$ to be the inf of $\{s_1,s_2,s_3\ldots\}$. Define $t_2$ to be the inf of $\{s_2,s_3,s_4\ldots\}$. $t_3$ to be the inf of $\{s_3,s_4,s_5\ldots\}$. and so on. Now check that $\{t_n\}$ is an increasing sequence, and therefore has a limit (possibly infinite). This limit is called the lower limit.