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I have to prove the following:

Given that lim $a_n$ exists and that lim $a_n$ = $a\in \mathbb{R}$ prove that: $$\limsup_{n \to \infty} (a_n+b_n)= a + \limsup_{n \to \infty} b_n$$

I proved one side of the inequality that is, I showed that $$\limsup_{n \to \infty} (a_n+b_n) \le a + \limsup_{n \to \infty} b_n$$ by using supremums of $(a_n)$, $(b_n)$ and $(c_n)$ where $(c_n)$={$c_k$|$k$$\ge$$n$} and $c_k=a_k+b_k$. But I can't figure out how to prove the converse. Can anyone explain how to do that without using $\epsilon$ or subsequences ? Any help would be appreciated thanks.

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  • $\begingroup$ Without using $\varepsilon$ or subsequences??? Then how? $\endgroup$
    – Berci
    Mar 30, 2015 at 21:47
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    $\begingroup$ Did you try to adapt the proof of any of the numerous near-duplicates listed in the column on the bottom right of the page? $\endgroup$
    – Did
    Mar 30, 2015 at 21:48
  • $\begingroup$ @Did Yes I did, but since most seem to involve subsequences or $\epsilon$ I wanted an answer which did not involve either of them something similar to what I did for one part of the inequality. $\endgroup$ Mar 30, 2015 at 21:56
  • $\begingroup$ You did but you say "most" and "seem"? $\endgroup$
    – Did
    Mar 30, 2015 at 21:58
  • $\begingroup$ @Did i didn't look at all of them as I'd already done those exact questions a while ago and wanted to look at only those which were either exactly the same or very similar to the question above. That is why I used "most" and "seem". I also looked for similar questions on the internet but they involved subsequences. Since I didn't think of a solution involving subsequences at first and did the proof another way I wanted to see if it was possible to do the converse following the same method. $\endgroup$ Mar 30, 2015 at 22:04

1 Answer 1

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Since

\begin{align}\limsup_{n\to \infty} (a_n + b_n) - \liminf_{n\to \infty} a_n &= \limsup_{n\to \infty} (a_n + b_n) + \limsup_{n\to \infty} (-a_n)\\ &= \lim_{n\to \infty} [\sup_{k\ge n} (a_k + b_k) + \sup_{k\ge n} (-a_k)]\\ &\ge \lim_{n\to \infty} \sup_{k\ge n}[(a_k + b_k) + (-a_k)] \\ &= \lim_{n\to \infty} \sup_{k\ge n} b_k\\ &= \limsup_{n\to \infty} b_n, \end{align}

we have

$$\limsup_{n\to \infty} (a_n + b_n) \ge \liminf_{n\to \infty} a_n + \limsup_{n\to \infty} b_n = a + \limsup_{n\to \infty} b_n.$$

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  • $\begingroup$ In this case, $a_k$ and $b_k$ are subsequences? $\endgroup$ Mar 30, 2015 at 22:07
  • $\begingroup$ Oh ok I got it. $\endgroup$ Mar 30, 2015 at 22:14

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