# Real Analysis - Can $\limsup(x_n)$ be defined using $<$ instead of $>$?

Let $$\{x_n\}_{n\in\infty}$$ be a sequence. Define for each $$n$$ the following subsets:

$$S_n := \{x_i \mid i\leq n\}$$ and $$\text{ }$$ $$T_n := \{x_i\mid i\geq n\}$$.

Now, using the $$\textit{supremum}$$ as a function on subsets, create the corresponding sequences:

$$\{\sup(S_n)\}_{n\in\mathbb{N}}$$ and $$\{sup(T_n)\}_{n\in\mathbb{N}}$$.

The question is, are the limits of these sequences the same? I.e. does

$$\lim\limits_{n\to\infty}\{\sup(S_n)\}_{n\in\mathbb{N}} \text{ }=?\text{ } \lim\limits_{n\to\infty}\{\sup(T_n)\}_{n\in\mathbb{N}} \text{ }\text{ }\left(=: \limsup\{x_n\}_{n\in\mathbb{N}}\right)$$

In english, this question is asking why we define limsup using the so called "tails" of the sequences instead of the "heads".

Computationally speaking, computing sup's for the heads and analyzing the trend would be more feasible than doing so for the infinite tails (as well as more intuitive from a graphical standpoint).

[Update: The "=?" does not hold as has been pointed out. The easy counter example being $$x_n := 1/n$$. It looks like monotone decreasing sequences would yield counter examples in general. The ones I tried were all increasing or oscillatory of constant amplitude (e.g. $$x_n := Cos(n*\pi/2)$$) or both. Thanks everyone!]

• There are obvious counter-examples. Did you try any? Commented Jul 14, 2021 at 23:13
• $n\in\infty$ is meaningless. What do you mean by it that is not included in $\lim_{n\to\infty}?$ Commented Jul 14, 2021 at 23:13
• The computational meaning of $\limsup$ is tricky. No finite amount of information about a sequence $(S_n)_{n=1}^\infty$ is sufficient to determine $\limsup_{n\to\infty}S_n$: the heads are irrelevant to the value of $\limsup$, the value of $\limsup$ is all about the behaviour in the limit of the tails. Commented Jul 15, 2021 at 2:04
• Or just consider the sequence $1000,\frac12,\frac23,\frac34,\frac45,\dots$. In any kind of limit, the influence of the early terms is supposed to fade away, with your proposed definition it doesn't.
– bof
Commented Jul 16, 2021 at 3:46

For example $$a_n := 1/n$$, take the $$\sup$$ of both sets given this sequence and see.