# Different definitions for $\limsup$

I am wondering if the following various definitions of $$\limsup$$ are equivalent:

$$\displaystyle{\limsup_{n \to \infty} x_n = \lim_{n \to \infty} \sup_{m \geq n} \{ x_m \}}$$ $$\displaystyle{\limsup_{n \to \infty} x_n = \lim_{n \to \infty} (\sup_{m \geq n} x_m)}$$ $$\displaystyle{\limsup_{n \to \infty} x_n = \inf_{n \geq 0} \sup_{m \geq n} x_m}$$

I pulled these from various sources and mostly I am interested in the different usages of the parentheses and how does one "read" these definitions step by step.

• do you know what is an accumulation point of a sequence? – janmarqz May 20 '20 at 19:08
• It should be $n\to \infty$ every time you have written $x\to\infty$. – Gae. S. May 20 '20 at 19:09
• Also, what's the difference between $\lim_{n\to\infty}\sup_{m\ge n}\{x_m\}$ and $\lim_{n\to\infty}\left(\sup_{m\ge n} x_m\right)$ ? – Gae. S. May 20 '20 at 19:10
• @Gae.S. Typo has been corrected. Regarding your last comment, I don't know, that is why I asked because I was confused with parentheses. So all of these are in fact the same? – Michael Munta May 21 '20 at 8:14
• take a look at math.stackexchange.com/a/3673558/568204 In this answer, I explain how to "read" the RHS. – peek-a-boo May 21 '20 at 8:22

The short answer is yes, they are all equivalent (assuming as a commenter mentioned that you replace $$x\longrightarrow \infty$$ with $$n\longrightarrow \infty$$).
Intuitively, limsup is a limit of a sequence, where the sequence's n-th term is given by sup$$_{m\geq n}x_m$$ (presuming we started out with some given initial sequence $$(x_0, x_1, x_2, x_3, …)$$ that is bounded). For instance, we get the 4-th term of the sequence by looking at the terms of the original sequence from $$x_4$$ onwards, and taking the supremum of those values. For the 5-th term, take the supremum of all terms of the original sequence from $$x_5$$ onwards, etc. Since successive terms of this new sequence are being formed by taking the supremum of progressively smaller subsets of the original sequence, we can conclude that this new sequence is (monotonically) decreasing. Thus it is also bounded above (the first term is the largest), and we know it is bounded below since the original sequence is bounded below. A sequence being bounded and decreasing is enough to infer that it converges, and so we can take its limit, and we call this value the limsup$$(x_n)$$ (since it is the limit of a sequence formed by taking the supremums of subsets of some original sequence $$(x_n)$$).
This fundamental intuition should show you that all of the definitions you gave are equivalent. The first two are trivially equivalent - the only difference is notational, in terms of where you prefer to place parentheses. But the last one is equivalent to the first two by the above. We noted that the supremum sequence was decreasing and bounded, and so it has a well-defined limit and infimum, and these values are the same. Therefore, $$\text{lim}_{n\longrightarrow\infty}\text{sup}_{m\geq n}x_m = \text{inf}_{n\geq 0}\text{sup}_{m\geq n}x_m$$.
Edit: On the left hand side, you wrote limsup$$_{n\longrightarrow\infty}(x_n)$$, but that doesn't really make sense (based on what I just wrote above). It is just limsup$$(x_n)$$. But still on the right-hand-side you do want to replace $$x\longrightarrow\infty$$ with $$n\longrightarrow\infty$$ for that value to make sense.