Question about the limsup of a sequence of real numbers Let $\{p_n\}$ be a sequence of real numbers. Is it true that if $p_n$ diverges, i.e. $p_n \rightarrow \infty$, then $\limsup~p_n=\infty$? Here $\limsup~p_n$ is the supremum of the set of all subsequential limits of $\{p_n\}$. My guess would be yes, $\limsup~p_n=\infty$, because the definition of subsequence does not preclude $\{p_n\}$ from being a subsequence of itself. Is this a correct line of reasoning? 
 A: Another definition of $\lim\sup$, which can be clarifying, is that $\lim\sup x_n$ is the supremum of $x$ such that $x_n>x$ for infinitely many $n$. I think the result is clearer from this perspective, as well as the fact that divergence is stronger than infinitude of the $\lim\sup$. In shorthand, a sequence diverges if it's eventually always arbitrarily large; but it has infinite $\lim\sup$ if it's arbitrarily large infinitely often.
A: You are correct that $\{p_{n}\}$ is a subsequence of $\{p_{n}\}$.  Thus, $\lim_{n \to \infty} \, sup \, p_{n}$ of a sequence $\{p_{n}\}$ that diverges to infinity is infinity.  Indeed, many divergent sequences $\{p_{n}\}$ have $\lim_{n \to \infty} \, sup \, p_{n}=\infty$.
Examples:
(a) If $p_{n}=n$, then $\lim_{n \to \infty} \, sup \, p_{n}=\infty$.
(b) If $\{p_{n}\}$ denotes the sequence $1, 1, \frac{1}{2}, 2, \frac{1}{3}, 3, ...$ , then $\lim_{n \to \infty} \, sup \, p_{n}=\infty$.
However, not every divergent sequence diverges to infinity.  Moreover, it is not true that $\lim_{n \to \infty} \, sup \, p_{n}=\infty$ for all divergent sequences $\{p_{n}\}$.  As a counterexample, consider the divergent sequence given by $p_{n}=i^{n}$.  We have $\lim_{n \to \infty} \, sup \, p_{n}= 1$.
