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Problem:

Let $\{a_n\}$ and $\{b_n\}$ be bounded sequences.

Prove $\{a_n + b_n\}$ is bounded and that $$\limsup_{n\to\infty} (a_n + b_n) \leq \limsup_{n\to\infty} a_n + \limsup_{n\to\infty} b_n$$

Theorems:

  1. If $\{a_n\}$ bounded then there exists subsequence $\{a_{n_k}\}$ converging to $\limsup a_n$

  2. If $\{a_n\}$ bounded, $\{a_{n_k}\}$ is subsequnce then $\limsup a_{n_k} \leq \limsup a_n$

Attempt:

Let there be a subsequence $\{a_{n_k} + b_{n_k}\}$ converging to $\limsup (a_n + b_n)$.

\begin{align}\limsup (a_n + b_n) &= \lim(a_{n_k} + b_{n_k})\tag{$\dagger$}\\ &= \lim(a_{n_k}) + \lim(b_{n_k})\tag{Property of limits}\\ &= \limsup (a_{n_k}) + \limsup (b_{n_k})\tag{Theorem 1}\\ &\leq \limsup (a_{n}) + \limsup (b_{n})\tag{Theorem 2}\end{align}

$(\dagger)$ - Subsequence converges to the same limit as the original sequence.

Thus proving the original statement.

Are there any flaws in my proof? If so how would I go about solving it?

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  • $\begingroup$ One issue is that $\lim a_n +b_n= \lim a_n + \lim b_n$ is only valid if $\lim a_n$ and $\lim b_n$ exist. $\endgroup$ – p.s. Mar 16 '17 at 8:58
  • $\begingroup$ @p.s. Can that be alleviated by using the Bolzano-Weirstrauss theorem in some way. I just learned about it so I'm unsure how to use it, this is just a guess since this problem did come from the same section that Bolzano-Weirstrauss was introduced. $\endgroup$ – Sky Mar 16 '17 at 9:17
  • $\begingroup$ Yes, by B-W you can take a subsequence of that subsequence such that the a's converge. $\endgroup$ – p.s. Mar 16 '17 at 16:58
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Yes, the application of Thm 1 is incorrect or unclear. It states that if $a_n$ is bounded there exists a subsequence that coverges to $\limsup$ - it does not follow that any subsequence has to converge to $\limsup$.

It's easy to fix this however. The step is basically correct, but you don't use Thm1, instead you're using that if the $\lim$ exists it will equal $\limsup$.

We can prove that since if $a_n$ is convergent any of it's subsequences must converge to the same limit, but there's a subsequence converging to $\limsup$ and that converges to the $\lim$ so $\lim=\limsup$.

Note that the result holds even if we drop the requirement that $a_n$ and $b_n$ are bounded (except if the RHS becomes undefined). The proof is basically the same, you just have to update Thm1 and Thm2 to reflect this.

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