$\lim \sup$ of a sequence

Question

Let $\{A_n\}$ be a sequence and $\frac{1}{R} = \lim \sup A_n$. Let $\alpha < R$. My question: Why is there $n_0\in \Bbb N $ such that $$A_n < \frac{1}{\alpha}\text{ for any } n\geq n_0$$ Thanks for your help.

Answer 1

One good definition of $\limsup_{n \to \infty} a_n = \sup E$ where the set $E$ is all limits of subsequences of the sequence $(a_n)$.

Now suppose $\frac{1}{R} = \limsup A_n$ and let $\alpha < R$. Then $\frac{1}{\alpha} > \frac{1}{R} = \limsup A_n$. Now $\epsilon > 0$, then I claim that there exists an $N \in \mathbb{N}$ such that $A_n < \frac{1}{R} + \epsilon$ for all $n \geq N$. Suppose otherwise, then there would exist infinitely many $n$ such that $A_n \geq \frac{1}{R} + \epsilon := \beta$ which means that there exists a subsequence $(A_{n_k})$ of $(A_n)$ so that $A_{n_k} \geq \beta$. Now we assume that $(A_k)$ is bounded above or else clearly $\limsup A_n = \infty$. So there exists an $M$ such that $M \geq A_{n_k} \geq \beta$ so since every bounded sequence has a convergent subsequence, we can conclude that there exists a subsequence $A_{m_k}$ that converges to a number $\phi \geq \beta > \limsup_n A_n$, which is a contradiction.

Answer 2

Let $R' = 1/R$ and $α' = 1/α$ to not confuse things too much. Rephrase your question:

Let $(A_n)_n$ be a real sequence and $R' = \limsup A_n$. Let $α' > R'$. Why is there $n_0 ∈ ℕ$ such that $A_n < α'$ for any $n ≥ n_0$?

Then the statement becomes a bit more obvious if you define/interpret $\limsup A_n$ to be the largest limit point of a sequence bounded from above. If there were infinitely many members $A_n$ above $α'$ in the statement, then there’d be a limit point greater than $α'$ (we assume the sequence is bounded from above), contradicting $α' > R' = \limsup A_n$.

In the case that $A_n$ is not bounded from above, then $\limsup A_n = ∞$, rendering the statement either meaningless or trivial.