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.

  • 7
    $\begingroup$ Hint: $$\limsup\limits_{n\to\infty}\; s_n = \limsup\limits_{n\to\infty}\; [(s_n+t_n)-t_n] \leq \cdots$$ and recall that $\limsup\limits_{n\to\infty}\;(-t_n) = -\liminf\limits_{n\to\infty}\;t_n.$ $\endgroup$ – t.b. Oct 7 '11 at 2:23
  • 1
    $\begingroup$ 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 $\limsup\limits_{n \rightarrow \infty}$ $s_n$. $\endgroup$ – emka Oct 7 '11 at 4:25
  • $\begingroup$ Related threads: math.stackexchange.com/questions/33888/… math.stackexchange.com/questions/69391/… $\endgroup$ – Martin Sleziak Oct 7 '11 at 7:05
  • $\begingroup$ 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. $\endgroup$ – t.b. Oct 7 '11 at 9:24
  • $\begingroup$ @t.b. But how to prove the identity $\lim\ \sup_{n\to \infty} (-t_{n})=-\lim\ \inf_{n\to \infty}\ t_n$ you mentioned? $\endgroup$ – Scorpio19891119 Jan 9 '13 at 0:10

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

  • $\begingroup$ Can you please explain it a bit further into details? Why does the inequality holds for all values of n but finitely number of n's? Shouldn't it hold by definition for an infinite number of n's? And moreover, why could you say that sn+tn>sn+t−εsn+tn>sn+t−ε results in lim sup(sn+tn)≥lim sup(sn+t−ε) $\endgroup$ – user371583 Nov 26 '16 at 17:05
  • $\begingroup$ @S.Peter 1) This is definition of limsup. (To be more precise, one of several definitions which are commonly used.) 2) If $a_n\le b_n$ for all $n$, then $\limsup a_n \le \limsup b_n$. (And if finitely many valued of a sequence are change, limit superior is not influenced by that.) $\endgroup$ – Martin Sleziak Nov 26 '16 at 17:10
  • $\begingroup$ You meant that this is the definition of lim inf in? From what I know the definition for both lim inf and lim sup is that there is an inifnitely number of n's that |a_n - L| < \$epsilon. Is your definition coming from this statement? I'm still confused regarding this issue $\endgroup$ – user371583 Nov 26 '16 at 17:34
  • $\begingroup$ @S.Peter See here for the definition of $\limsup$. The limit inferior is defined similarly. (As I mentioned, several definitions are commonly used, all of them are equivalent. But this is what is used in the first part of the post.) $\endgroup$ – Martin Sleziak Nov 26 '16 at 17:57
  • $\begingroup$ Okay I understand the definition part. But regarding the (And if finitely many valued of a sequence are change, limit superior is not influenced by that) that you mentioned, isn't it true only when the related sequence is converged? In this question it is possible that it isn't, am I worng? $\endgroup$ – user371583 Nov 26 '16 at 19:09

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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