# Proof of theorem “If $\lim s_n$ is defined, then $\liminf s_n=\lim s_n=\limsup s_n$”

I am reading Ross textbook Elementary analysis the theory of calculus and in a theorem, we are trying to prove the statement

If $\lim s_n$ is defined, then $\liminf s_n=\lim s_n=\limsup s_n$.

I am having troubles understanding the case when $\lim s_n = s$. The proof says:

Suppose that $\lim s_n=s$, where s is a real number. Consider $\epsilon >0$. There exists a natural number N such that $|s_n-s|<\epsilon$ for $n>N$. Thus $s_n<s+\epsilon$ for $n>N$, so $$v_N=\sup{\{s_n:n>N\}} \leq s+\epsilon$$ where $v_N=sup{\{s_n:n>N\}}$.

The next line says that "Also, $m>N$ implies $v_m \leq s+\epsilon$, so $\limsup s_n=\lim v_m \leq s+ \epsilon$. This is the line I don't get, why is it that from the statement $v_m \leq s+ \epsilon$, we got to the statement $\limsup s_n = \lim v_m\leq s+\epsilon$? Did we just took the limit of both sides of the inequality and since the LHS is a sum of constants, it stays the same?

• Do you recall the definitions of limsup and liminf? 'Cause, from what you explain, this seems to be the missing piece... – Did Sep 21 '15 at 0:26

Let $\epsilon>0$ be the one given in your proof. Then there exists some natural $N$ such that $v_n\leqslant s+\epsilon/2$ for all $n\geqslant N$ (this is the part you do understand).
Let $l:=\lim\limits_{n\to\infty}v_n$ and let $M$ be such that $(|v_n-l|<\epsilon/2\;\;\Longrightarrow\;\;l<v_n+\epsilon/2)$ whenever $n\geqslant M.$ It follows that if $n\geqslant\max\{N,M\}$ then $l<v_n+\epsilon/2\leqslant(s+\epsilon/2)+\epsilon/2=s+\epsilon$ and hence $l<s+\epsilon.$
Note that since $\epsilon$ is arbitrary, we can conclude that $l\leqslant s.$