Proof correctness of: $\limsup a_n = \infty \implies \exists{a_{n_k}}$ such that $a_{n_k} \to \infty$

Question

Prove: $\limsup a_n = \infty \implies \exists{a_{n_k}} (a_{n_k} \to \infty)$

Is this correct?

Proof:

Consider $a_n^* =\inf\{a_k : k \geq n\}$. Thus, $a_n^*$ is monotone non-decreasing. $\limsup a_n = \infty \implies a_n^*$ is unbounded.

Thus, $a_n^*$ is unbounded and montone non-decreasing.

Therefore $a_n^* \to \infty$

Let me know if I need stronger justifications or how I can improve the clarity of my proof.

Answer 1

First of all, you should use the fact that the sequence $$a_k^*=\sup\{a_n; n\ge k\}$$ converges to $+\infty$. (Since $\limsup a_n =\lim_{k\to\infty} \sup\{a_n; n\ge k\}$.) The same is not true about $\inf\{a_n; n\ge k\}$.

What we know about the sequence $a_k^*$.

• It is non-increasing: $a_k^*\ge a_{k+1}^*$.
• Since it is non-increasing, we know that $\lim_{k\to\infty} a_k^* = \inf_k a_k^*$. * The equation $\inf_k a_k^*=\infty$ implies that each $a_k^*$ is equal to $+\infty$.

If supremum of some set of real numbers is $+\infty$, then it contains arbitrarily large numbers. So we can construct $a_{n_k}$ inductively as follows: If we know $a_{n_1},\dots,a_{n_k}$, then we choose $a_{n_{k+1}}$ in a such way that:

• $n_{k+1} > n_k$;
• $a_{n_{k+1}} \ge k+1$.

We know existence of such $n_{k+1}$ from the fact that $\sup\{ a_n; n>n_k\}=+\infty$.

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Maybe it is worth mentioning that this result is true not only for $+\infty$. If $S=\limsup a_n$, then there is a subsequence $(a_{n_k})$ such that $S=\lim_{k\to\infty} a_{n_k}$.

Answer 2

I would suggest to define $n_k=\inf\{t\in\mathbf{N},a_t>k\}$. For each $k$, the set $\{a_t>k\}$ is non-empty because of your assumption on limsup. The sequence $(n_k)_k$ is increasing.

The sequence $(a_{n_k})_k$ is a subsequence of $(a_n)$ and its limit is $+\infty$.

In the original post, your $(a_n^*)_n$ might be constant (define $a_{2n}=1$ and $a_{2n+1}=n$).