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$$\limsup_{n\to\infty} (a_n+b_n)\leq \limsup_{n\to\infty} a_n + \limsup_{n\to\infty} b_n$$

I know there's a proof somewhere on here but I just can't seem to understand it. Can anyone give me an exemple why this is true. In my head, I always see it as equal... it's like I take the supremum from the set $A$ and I take an supremum of the set $B$ and I add it up... where I'm I wrong?

Thank you for your support.

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    $\begingroup$ Here's you can find smth useful math.stackexchange.com/questions/1812070/… $\endgroup$ – ZFR Jun 11 '16 at 16:47
  • $\begingroup$ a_n+b_n<= limsup(a_n)+limsup(b_n) . limsup {a_n+b_n}<=limsup {(limsup(a_n)+limsup(b_n)}=limsupa_n+limsupb_n $\endgroup$ – Jacob Wakem Jun 11 '16 at 22:20
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Suppose $a_n = (-1)^n$ and $b_n = (-1)^{n+1}$, so we have this: \begin{align} (a_n : n=1,2,3,\ldots) & = \left( \begin{array}{rrrrrrrr} -1, & 1, & -1, & 1, & -1, & 1, & \ldots \end{array} \right) \\ (b_n : n=1,2,3,\ldots) & = \left( \begin{array}{rrrrrrrr} 1, & -1, & 1, & -1, & 1, & -1, & \ldots \end{array} \right) \end{align} Then $a_n+b_n=0$ for all $n$, so $\limsup\limits_{n\to\infty} (a_n+b_n) = 0$.

But $\limsup\limits_{n\to\infty} a_n + \limsup\limits_{n\to\infty} b_n = 1+1=2$.

That shows how the two sides can fail to be equal. Why the opposite inequality cannot hold is another question; maybe I'll come back and add that later.

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  • $\begingroup$ when $a$ and $b$ are correlated, it's possible to get less $\endgroup$ – cactus314 Jun 11 '16 at 17:43
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Michael Hardy gave an example when the two sides are not equal. I will provide a proof of the inequality in the OP. First, we recall the definition of the limit superior:

$$ \limsup_{n\to \infty} a_n = \lim_{n\to \infty} \sup_{m \geq n} a_m = \inf_{n \in \mathbb{N}} \sup_{m \geq n} a_m $$

Now, we prove the statement. We have that

$$ \limsup_{n\to \infty} a_n + b_n = \lim_{n\to \infty} \sup_{m \geq n} a_m + b_m \leq \lim_{n\to \infty} \left(\sup_{m \geq n} a_m + \sup_{m \geq n} b_m \right) = \lim_{n\to \infty} \sup_{m \geq n} a_m + \lim_{n\to \infty} \sup_{m \geq n} b_m = \limsup_{n\to \infty} a_n + \limsup_{n\to \infty} b_n $$

The key inequality is $ \sup_{m\geq n} (a_m + b_m) \leq \sup_{m\geq n} a_m + \sup_{m\geq n} b_m $. This inequality holds because the right hand side is clearly an upper bound for $ a_m + b_m $, therefore it is greater than or equal to the LHS by definition of the least upper bound.

Note that this also tells us when equality might fail. For another example, you may consider the following:

$$ \limsup_{x\to \infty} (\sin(x) + \cos(x)) = \sqrt{2} $$ $$ \limsup_{x\to \infty} \sin(x) + \limsup_{x\to \infty} \cos(x) = 2 $$

The problem, as stated by Bernard, is that $ \sin(x) $ and $ \cos(x) $ do not attain their supremum values at the same points.

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  • $\begingroup$ Your proof is predicated on the statement "This inequality holds because the right hand side is clearly an upper bound ..." I suggest considering expanding on this. ;-)) -Mark $\endgroup$ – Mark Viola Jun 11 '16 at 22:16
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It is not like taking the supremum of $A$ and the supremum of $B$ and adding them to obtain the supremum of $A+B$, because the values in $A$ and $B$ are indexed by a natural number $n$, and the supremum for $A$ and $B$ are not necessarily attained at the same index $n$. Hence the sum of the suprema is only an upper bound for the sum of elements with the same index.

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