I'm trying to show that if $ \limsup s_n = +\infty$ and $\liminf t_n > 0$, then $\lim\sup s_n t_n = +\infty$.

Could someone check my proof/give feedback?

Since $\lim\inf t_n > 0$, we know that there is a natural $N_1$ such that $m = \inf \{t_n \ | \ n > N_1 \} > 0$. Also because $\lim\sup s_n = +\infty$, there exists $N_2 \in \mathbb{N}$ and $M > 0 $ such that $\displaystyle \sup s_n > \frac{M}{m}$ for $n > N_2$. Then for $n > \max \{ N_1, N_2 \}$, we have $s_n t_n \geq s_n m$. Now we can write $\displaystyle \lim\sup s_n t_n \geq \lim\sup s_n m = m \lim\sup s_n > m \left( \frac{M}{m} \right) = M$. Since $M$ was any arbitrary number greater than 0, we see that $\lim\sup s_n t_n = +\infty$.


Since $\limsup s_n = +\infty$, there exists a subsequence $s_{n_j} \to +\infty$ as $j \to \infty$. Then we can also take a subsequence $t_{n_j}$ of $t_n$. Now we have $s_{n_j}t_{n_j} \geq s_{n_j}m$ and $\limsup s_{n_j}t_{n_j} \geq m \limsup s_{n_j} = +\infty$. Since $\limsup s_{n_j}t_{n_j} = +\infty$, we must have $\limsup s_n t_n = +\infty$

  • $\begingroup$ $s_{n_j}t_{n_j} \geq s_{n_j}m$ for $n_j >N_1$. Moreover, you have better than $\limsup s_{n_j}t_{n_j}=+\infty$; you actually have $\lim_{n\to\infty}s_{n_j}t_{n_j}=+\infty$. This lets you make the final conclusion. $\endgroup$ – Gyu Eun Lee May 21 '13 at 5:12
  • $\begingroup$ @proximal: Awesome! Thanks for the followup. $\endgroup$ – Student May 21 '13 at 14:07

This is red herring, but there are a few mistakes in this line:

"...because $\limsup s_n=+\infty$, there exists $N_2\in\mathbb{N}$ and $M>0$ such that $\sup s_n>\frac{M}{m}$ for $n>N_2$."

First of all, if by $\sup s_n$ you mean $\sup\{s_n: n \in \mathbb{N}\}$ then clearly this number is $\infty$, so your statement is vacuous. If by $\sup s_n$ you are referring to $\sup\{s_n:n>N_2\}$, then this number is still $+\infty$.

Second, this $N_2$ will depend on $M$; the correct order of quantification is "for any $M>0$ there exists $N_2\in\mathbb{N}$ such that..."

I say this is a red herring because this line is irrelevant to the proof. You are done once you are able to show that $\limsup s_nt_n \geq m\limsup s_n$, since the right hand side is equal to $+\infty$. You have not quite done this yet. Instead of doing this strange business with the $\limsup$, try going back to the original definition:

By $\limsup a_n=+\infty$ we mean that there exists a subsequence $(a_{n_k})_{k=1}^\infty$ such that $a_{n_k}\to+\infty$ as $k \to \infty$.

Getting the inequality $s_nt_n\geq s_nm$ is a good idea (in fact, this is the only hard part of the proof), but you need to ensure $s_n \geq 0$ for this inequality to work. Since you're working with $\limsup$, you don't need the entire sequence to be nonnegative, but only a subsequence. Can you extract such a subsequence from the definition?

  • $\begingroup$ Thanks for the feedback. Would you mind looking at my edit? $\endgroup$ – Student May 21 '13 at 4:09

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