How to prove that the limsup of a sequence is equal to its greatest subsequential limit?

Solution 1:

Use the definition: $\lim \sup S_n=\lim_{n\to \infty}T_n$ where $T_n=\sup \{S_m:m\geq n\}.$ Assume $-\infty< L=\lim \sup S_n<+\infty.$

(I). Take any $M>L.$ There are only finitely many $n$ such that $S_n\geq (M+L)/2.$ Because otherwise $T_n\geq (M+L)/2$ for every $n,$ implying $\lim_{n\to \infty}T_n\geq (M+L)/2>L.$

So $(S_n)_n$ cannot have a subsequence converging to $M$ for any $M>L.$ Because otherwise there are infinitely many $n$ with $|S_n-M|<(M-L)/2,$ implying there are infinitely many $n$ such that $S_n\geq (M+L)/2.$

(II). Take any $N<L.$ There do exist infinitely many $n$ such that $S_n>N.$ Otherwise $T_n\leq N$ except for finitely many $n,$ implying $\lim_{n\to \infty}T_n\leq N<L.$

(III). By (II) with $N =L-2^{-1},$ take $n(1)$ such that $S_{n(1)}> L-2^{-1}.$ Recursively, by (II) with $N=L-2^{-j-1}$ take $n(j+1)>n(j)$ such that $S_{n(j+1)}> L-2^{-j-1}.$

The sequence $(S_{n(j)})_j$ converges to $L.$

Because if $\epsilon >0,$ then by (I) with $M=L+\epsilon$, take $K\in \mathbb N$ large enough that $n\geq K\implies S_n<L+\epsilon.$ And take $j_K\in \mathbb N$ large enough that $n(j_K)\geq K$ and $2^{-j_K}<\epsilon.$ Then $$j\geq j_K\implies |S_{n(j)}-L|<\epsilon.$$

Remark: $\lim \sup S_n$ is the the largest number $x$ such that, for every $r>0,$ the set $\{n: S_n\in [-r+x,r+x]\;\}$ is infinite.

Solution 2:

You have to prove it in two steps.

First, prove that the greatest sequential limit cannot be greater than the limsup (that's easy, using reductio ab absurdum, suppose a sequential limit is greater than the limsup, and derive a contradiction).

Then, prove that the greatest sequential limit cannot be smaller than the limsup (explicitly build a sequence whose limit is greater than any real number strictly smaller than the limsup).

The only other option is that those two limits are equal.