I am stuck with the following problem.

Prove that $$\limsup_{n \to \infty} (a_n+b_n) \le \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n$$

I was thinking of using the triangle inequality saying $$|a_n + b_n| \le |a_n| + |b_n|$$ but the problem is not about absolute values of the sequence.

Intuitively it's clear that this is true because $a_n$ and $b_n$ can "reduce each others magnitude" if they have opposite signs, but I cannot express that algebraically...

Can someone help me out ?

  • $\begingroup$ I'm sorry, my book (Rudin) doesn't use the notation $\sup_{k \ge n}a_k$ and I am not sure what it means. $\endgroup$ – hyg17 Jun 1 '13 at 2:11

Define for all natural numbers $k$: $A_k = \sup\{ a_n: n \ge k \}$, $B_k = \sup\{ b_n: n \ge k \}$ (where $A_k, B_k \in \mathbb{R} \cup \{+\infty\}$, (They are decreasing because for larger $k$ we take the $\sup$ of fewer terms), so that by definition $\limsup_{n \to \infty} a_n = \lim_{k \to \infty} A_k$ and similarly for $B_k$ and $\limsup_{n \to \infty} b_n$. Also we consider the $C_k = \sup \{ (a_n + b_n) : n \ge k \}$, so that $\lim_{k \to \infty} C_k = \limsup_{n \to \infty} (a_n+b_n)$.

Now, fix an index $k$, then for all $n \ge k$ we have $a_n + b_n \le A_k + B_k$, because we estimate $a_n$ by the supremum of all terms of $(a_n)$ with $n \ge k$ and likewise for the $b_n$. As (for fixed $k$) the right hand side is fixed:

$$C_k = \sup \{ (a_n + b_n : n \ge k \} \le A_k + B_k\mbox{.}$$

This holds for all $k$, so we take the $\inf$ or $\lim$ on both sides as $k$ tends to infinity, and this preserves the inequality and we are done.

  • $\begingroup$ It's looking a lot clearer, but I'm still puzzled about the part where you said, "$a_k+b_k \le A_k +B_k \Rightarrow a_k+b_k \le A_n+B_n$". If $A_n+B_n$ is smaller than $A_k+B_k$, how is the above inequality guaranteed ? $\endgroup$ – hyg17 Jun 1 '13 at 17:56
  • $\begingroup$ @hyg17 you are right, I fixed the argument (hopefully). $\endgroup$ – Henno Brandsma Jun 1 '13 at 18:19
  • $\begingroup$ Okay, now it's 100% lucid. Thanks ! It was a lot more helpful than you can imagine. $\endgroup$ – hyg17 Jun 1 '13 at 18:29
  • $\begingroup$ @hyg17 Glad to be able to help! $\endgroup$ – Henno Brandsma Jun 1 '13 at 18:31
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    $\begingroup$ @AshokVardhan try two sequences that alternate between two values $\endgroup$ – Henno Brandsma Aug 28 '15 at 15:34


Given two sequences, $\;\displaystyle \{a_n\}_{n \in \Bbb N},\;$ $\,\{b_n\}_{n \in \Bbb N},\;$

and given the definition of the supremum of a sequence, we can see that for every $k\geq n$, $$(a_k + b_k) \;\; \leq \;\;\sup_{k\geq n} a_k + \sup_{k\geq n} b_k\,.$$

Now how does this imply that $$\lim_{n\to \infty} \sup(a_n + b_n) \;\; \leq \;\; \lim _{n \to \infty} \sup a_n + \lim_{n\to \infty} \sup b_n\quad ?$$

Added: see Definition 3.16, Theorem 3.17, 3.19: perhaps you'd prefer to use the notation used there.

  • $\begingroup$ I am puzzled. From my understanding $\sup a_k$ and $\sup b_k$ is a fixed number, so if $a_k + b_k$ happens to be increasing, the supremum of $a_k+b_k$ can only equal to the sum of the suprema... right ? $\endgroup$ – hyg17 Jun 1 '13 at 2:10
  • $\begingroup$ Recall that we are taking the limit of the suprema as $n\to \infty$. The supremum say of $a_n$ is fixed for a fixed n, or for a fixed k. We are simply stating that lim sup of the sum of the terms (an + bn) is necessarily less than or equal to the lim sup of an + lim sup of bn, when $n \to \infty$. $\endgroup$ – Namaste Jun 1 '13 at 2:20
  • $\begingroup$ Examine Thm. 3.19 $\endgroup$ – Jake Color Jun 1 '13 at 2:20
  • $\begingroup$ @amWhy I don't understand. The supremum of a sequence is not always larger than the terms of a sequence. As a matter of fact, {$1 \over n$} is a sequence where all of it's temrs are larger than the supremum. So how can $a_k \le \sup a_k$ at all ? $\endgroup$ – hyg17 Jun 1 '13 at 2:31
  • $\begingroup$ @hyg17 for your example $(a_n) = (\frac{1}{n})$, $s_k = \sup \{a_n: n \ge k \} = \frac{1}{k}$ for all $k$ and its limit is $0$, which is thus the $\limsup a_n$. But the supremum of the tail of the sequence $s_k$ is at least as large as all the terms it is supremum of.. $\endgroup$ – Henno Brandsma Jun 1 '13 at 5:04

I happened to making solutions for PMA, so I would like to share my solution here:

If either $\limsup_{n \to \infty} a_n = +\infty$ or $\limsup_{n \to \infty} b_n = +\infty$, there is nothing to prove. So we may assume $\limsup_{n \to \infty} a_n = A, \limsup_{n \to \infty} b_n = B$, where $A < +\infty, B < +\infty$ (but each of them can possibly take $-\infty$).

Given $\varepsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$, such that $a_n < A + \varepsilon/2$ for all $n \geq N_1$ and $b_n < B + \varepsilon/2$ for all $n \geq N_2$. Take $N = \max(N_1, N_2)$, it follows that for all $n \geq N$, $$a_n + b_n < A + \varepsilon/2 + B + \varepsilon/ 2 = A + B + \varepsilon.$$ Let $n \to \infty$ in the above equation, we conclude that $\limsup_{n \to \infty} (a_n + b_n) \leq A + B + \varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $\limsup_{n \to \infty}(a_n + b_n) \leq A + B$, proving the result.


This question occurs in Rudin's PMA, page 78, Exercise 5. Below is my trial of proof. Is it right? Any comments are welcome!

5. $\quad $For any two real sequences $\{a_n\}, \{b_n\},$ prove that \begin{gather*} \limsup_{n\to\infty}(a_n+b_n)\leq \limsup_{n\to\infty} a_n+\limsup_{n\to\infty} b_n, \end{gather*} provided the sum on the right is not of the form $\infty-\infty.$

proof: $\quad$ Put $a^*=\limsup\limits_{n\to\infty} a_n, b^*=\limsup\limits_{n\to\infty} b_n$ and $c^*=\limsup\limits_{n\to\infty} (a_n+b_n).$ We shall show $c^*\leq a^*+b^*.$

Since it is excluded that the right is of the form $\infty-\infty,$ if one of $a^*$ and $ b^*$ is $-\infty,$ then the assertion holds. So we assume that $a^*, b^*>-\infty.$ And if one of $a^*$ or $b^*$ is $+\infty,$ then $c^*=a^*+b^*,$ so we we need to consider the case $a^*, b^*\in\mathbb{R}.$

By Theorem 3.17 (see Page 56 of Rudin's PMA), for every $\epsilon>0$ there exists $N_1, N_2$ such that \begin{gather*} a_n<a^*+\frac{\epsilon}{2},\qquad \forall n>N_1,\\ b_n<b^*+\frac{\epsilon}{2},\qquad \forall n>N_2. \end{gather*} Hence we have \begin{gather*} a_n+b_n<a^*+b^*+\epsilon,\qquad \forall n>\max\{N_1, N_2\}. \end{gather*} Then, for every convergent subsequence $a_{\sigma(n)}+b_{\sigma(n)}$ of sequence $(a_n+b_n)_{n\in\mathbb{N}},$ we have \begin{gather*} \lim_{n\to\infty} (a_{\sigma(n)}+b_{\sigma(n)})\leq a^*+b^*+\epsilon. \end{gather*} Thus we see that $a^*+b^*+\epsilon$ is an upper bound of the set $C$ of subsequential limits of $(a_n+b_n)_{n\in\mathbb{N}}.$ Hence $c^*=\sup C\leq a^*+b^*+\epsilon.$ By the arbitrariness of $\epsilon>0,$ it follows that $c^*\leq a^*+b^*.$ $\Box$


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