# Limsup(a+b) = limsup(a) + limsup(b)

## Main Question or Discussion Point

Let $\{a_n\}$ and $\{b_n\}$ be bounded sequences. Say that we already know that $\displaystyle \limsup_{n\to\infty} (a_n+b_n) \le \limsup_{n\to\infty}a_n + \limsup_{n\to\infty}b_n$.

But isn't it also true then that $$\limsup_{n\to\infty} b_n = \limsup_{n\to\infty} ((a_n+b_n) +(- a_n)) \le \limsup_{n\to\infty} (a_n+b_n) + \limsup_{n\to\infty} (-a_n) = \limsup_{n\to\infty} (a_n+b_n) - \limsup_{n\to\infty} a_n,$$ and so $\limsup_{n\to\infty} b_n + \limsup_{n\to\infty} a_n \le \limsup_{n\to\infty} (a_n+b_n)$. So we conclude that $\displaystyle \limsup_{n\to\infty} (a_n+b_n) = \limsup_{n\to\infty}a_n + \limsup_{n\to\infty}b_n$. Is something going wrong with this argument? I think in general you need one of the sequences to be convergent for this to be true.

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Math_QED
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$\limsup_{n \to \infty} (-a_n) \neq - \limsup_{n \to \infty} a_n$

Take $a_n = (-1)^n$.

$\limsup_{n \to \infty} (-a_n) \neq - \limsup_{n \to \infty} a_n$

Take $a_n = (-1)^n$.
Ah, I see. But if $a_n$ converges then $\limsup_{n \to \infty} (-a_n) = - \limsup_{n \to \infty} a_n$.

Math_QED
Ah, I see. But if $a_n$ converges then $\limsup_{n \to \infty} (-a_n) = - \limsup_{n \to \infty} a_n$.