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Let $\{s_n\}$ and $\{t_n\}$ be sequences.

I've noticed this inequality in a few analysis textbooks that I have come across, so I've started to think this can't be a typo:

$\limsup\limits_{n \rightarrow \infty} s_n + \liminf\limits_{n \rightarrow \infty} t_n \leq \limsup\limits_{n \rightarrow \infty} (s_n + t_n) \leq \limsup\limits_{n \rightarrow \infty} s_n + \limsup\limits_{n \rightarrow \infty} t_n$.

I understand the right two-thirds of the equation, and can prove it. I'm stuck on the far-left third. I don't understand how that is true.

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Hint: $$\limsup\limits_{n\to\infty}\; s_n = \limsup\limits_{n\to\infty}\; [(s_n+t_n)-t_n] \leq \cdots$$ and recall that $\limsup\limits_{n\to\infty}\;(-t_n) = -\liminf\limits_{n\to\infty}\;t_n.$ – t.b. Oct 7 '11 at 2:23
Is this leading towards something like: $\limsup\limits_{n \rightarrow \infty}$ $s_n = \limsup\limits_{n \rightarrow \infty}$ $[(s_n+t_n)-t_n] \leq \limsup\limits_{n \rightarrow \infty}(s_n+t_n)-\liminf\limits_{n \rightarrow \infty}$ $t_n$. Then I sub in the last part of the inequality for $\limsup\limits_{n \rightarrow \infty}$ $s_n$. – emka Oct 7 '11 at 4:25
Yes exactly, that was the intention. This reduces the claim to what you could prove. Alternatively, you could of course do as Martin did in his answer below. – t.b. Oct 7 '11 at 9:24
@t.b. But how to prove the identity $\lim\ \sup_{n\to \infty} (-t_{n})=-\lim\ \inf_{n\to \infty}\ t_n$ you mentioned? – Scorpio19891119 Jan 9 '13 at 0:10
up vote 7 down vote accepted

I will write only about the first inequality $$\limsup\limits_{n \rightarrow \infty} s_n + \liminf\limits_{n \rightarrow \infty} t_n \leq \limsup\limits_{n \rightarrow \infty} (s_n + t_n).$$ (You wrote that you can prove the rest.)

By definition, if $\liminf t_n=t$ then for each $\varepsilon>0$ the inequality $t_n>t-\varepsilon$ holds for all but finitely many $n$'s. For such $n$'s we also have $s_n+t_n>s_n+t-\varepsilon$ and $$\limsup(s_n+t_n) \ge \limsup (s_n+t-\varepsilon) = t-\varepsilon+ \limsup (s_n).$$ (We have used monotonicity of $\limsup$ and that $\limsup (C+x_n)=C+\limsup x_n$ for any constant $C$.) Since the above inequality is true for each $\varepsilon>0$, we get that $$\limsup(s_n+t_n) \ge t + \limsup (s_n)=\liminf t_n+\limsup s_n.$$

EDIT: Note that the above proof does not work for $t=-\infty$ (it does not make sense to write $-\infty-\varepsilon$), but in this case the inequality is clear. (Of course, we have to omit indeterminate case $\infty-\infty$, i.e., in this case we assume that $\limsup s_n$ is finite.)

Or if you use $\liminf\limits_{n\to\infty} x_n= \lim\limits_{n\to\infty} \inf\limits_{n\ge n_0} x_n$ and $\limsup\limits_{n\to\infty} x_n= \lim\limits_{n\to\infty} \sup\limits_{n\ge n_0} x_n$ as the definition of limit inferior/superior then you can use $$\sup_{n\ge n_0} (x_n+y_n) \ge \sup_{n\ge n_0} x_n + \inf_{n\ge n_0} y_n$$ to get $$\lim_{n\to\infty}\sup_{n\ge n_0} (x_n+y_n) \ge \lim_{n\to\infty}\sup_{n\ge n_0} x_n + \lim_{n\to\infty}\inf_{n\ge n_0} y_n\\ \limsup_{n\to\infty} (x_n+y_n) \ge \limsup_{n\to\infty} x_n + \liminf_{n\to\infty} y_n.$$

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This is an important result, so I would guess that most books and courses will probably mention it and either give a solution or left it as an exercises.

I'll just give a few links (I will not attempt to write down again what others have already written down - and probably much better - and what can be easily found).

EDIT: I apologize for not reading your question thoroughly enough. Most of the links I've given bellow prove only the second inequality. (Which is the easier one and you wrote that you were able to show it yourself.) I've left the remaining links - they might be useful for someone - but I've marked which of them show both inequalities.



  • Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series, Problem 2.4.15. The problem is given on p.44 and solved on p.198-199. (AFAIK this book is also available in French and Polish.) Both inequalities are shown here. You might also be interested in Problem 2.4.17, where similar result for product of sequences is shown.

  • Sterling K. Berberian: A first course in real analysis, p.54. One of the inequalities is given here as an exercise, but a detailed hint is given.

  • Jacques Dixmier: General Topology, Theorem 7.3.7 - this result is given here in a much greater generality - for a limit superior along a filterbase.


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First, the limit superior terms all cancel so really what you have written is equivalent to

$\liminf\limits_{n \rightarrow \infty}$ $t_n$ $\leq$ $t_n$ $\leq \limsup\limits_{n \rightarrow \infty}$ $t_n$

Consider $t_n = \frac{(-1)^n}{n}$.

$\limsup\limits_{n \rightarrow \infty}$ $t_n$ $=$ $\liminf\limits_{n \rightarrow \infty}$ $t_n$ $= 0$

However for every $n\in \mathbb{N}$, your theorem would imply $0\leq t_n \leq 0$ which means $t_n = 0$. $t_n \neq 0 \forall n\in\mathbb{N}$ so this would not even be true if you added a condition like "$\exists N\in \mathbb{N}$ such that this is true for all $n>N$".

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I think the intention was $$\limsup_{n\to\infty}\; s_n + \liminf_{n\to\infty}\; t_n \leq \limsup_{n \to \infty}\;{(s_n+t_n)}$$ – t.b. Oct 7 '11 at 2:01

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