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I'm using the definition where $ \lim \sup s_n = \lim ( \sup \{s_k : k \geq n \} )$.

I know $\lim \sup (s_n + t_n) \leq \lim \sup (s_n) + \lim \sup (t_n)$ makes intuitive sense because I'm guessing the sequence $s_n + t_n$ is much larger than than the individual sequences $s_n$ and $t_n$. And so the least upper bound would be larger.

However I'm stuck on proving this. I would know how to do this question if I could make the assumption that $s_n, t_n$ were bounded sequences, but I cannot make this assumption.

Martin Sleziak
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Nikitau
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1 Answers1

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The original statement (edited: concerning $\geq$ by a slip of the pen of the OP) is not true.

Let it be that $s_n=1$ if $n$ is odd and $s_n=0$ if $n$ is even.

Let it be that $t_n=1$ if $n$ is even and $t_n=0$ if $n$ is odd.

Observe that $s_n+t_n=1$ for every $n$.

Then: $$\limsup s_n+\limsup t_n=1+1=2>1=\limsup (s_n+t_n)$$

edit in order to prove that $\leq$ holds:

Fix $n$.

For every $l\geq n$ we have: $$s_{l}\leq\sup_{k\geq n}s_k\wedge t_{l}\leq\sup_{k\geq n}t_k$$ Consequently for every $l\geq n$ we have: $$s_l+t_l\leq\sup_{k\geq n}s_k+\sup_{k\geq n}t_k$$

That allows us to conclude for every $n$:$$\sup_{k\geq n}(s_k+t_k)\leq\sup_{k\geq n}s_k+\sup_{k\geq n}t_k$$ Then consequently:$$\lim_{n\to\infty}\sup_{k\geq n}(s_k+t_k)\leq\lim_{n\to\infty}(\sup_{k\geq n}s_k+\sup_{k\geq n}t_k)=\lim_{n\to\infty}\sup_{k\geq n}s_k+\lim_{n\to\infty}\sup_{k\geq n}t_k$$

drhab
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