I was trying to prove the following inequality

$$\limsup (a_n + b_n) \leq \limsup (a_n) + \limsup (b_n)$$

for the case when both are Real Numbers.

Let $a_{{n}_k}$ and $b_{{n}_k}$ be subsequences converging to limits $a_1$ and $b_1$, then given any $\epsilon > 0$, there exists $N_1$ and $N_2$ such that

$| a_{{n}_k} - a_1| < \epsilon /2\qquad\forall k>N_1$, and

$| b_{{n}_k} - b_1| < \epsilon /2\qquad\forall k>N_2$

and by triangle inequality we have

$$| a_{{n}_k} + b_{{n}_k} - (a_1+b_1) | < \epsilon\qquad\forall k>\max\{N_1,N_2\}$$

And since we know there exists subsequences converging to $\limsup a_n$ and $\limsup b_n$, we have $(a_{{n}_k} + b_{{n}_k}) \rightarrow\limsup a_n+\limsup b_n$. Something seems to be wrong, can anyone help me with this, Thank you.

  • 1
    $\begingroup$ The problem is about the indices…as you noticed somethings wrong, one counterexample is $a_n=(-1)^n$, $b_n =a_{n+1}$. $\endgroup$ Jun 17, 2017 at 8:45
  • $\begingroup$ I am sorry but I didn't quite understand. "The problem is about indices" $\endgroup$ Jun 17, 2017 at 8:48
  • $\begingroup$ The correct claim in last paragraph is: there are increasing sequences of natural numbers $(n_k)$ and $(m_k)$ such that $(a_{n_k}+b_{m_k}) \rightarrow …$ you have made a wrong statement there. $\endgroup$ Jun 17, 2017 at 8:54
  • $\begingroup$ You must add more arguments, like $a_n ,b_n$ are non-negative sequences $\endgroup$
    – Red shoes
    Jun 17, 2017 at 9:15
  • $\begingroup$ I don't think that would change anything, consider sequences $3+ (-1)^n~ and ~ 3+ (-1)^{n+1}$ $\endgroup$ Jun 17, 2017 at 9:22

3 Answers 3


The problem in your proof, as mentioned by @Lee Chun Min, is that you claimed that there was a subsequence of $c_n=a_n+b_n$ converging to $\limsup a_n+\limsup b_n$ from the fact that there exists subsequences $a_{n_k}$ and $b_{N_k}$ converging to $\limsup a_n$ and $\limsup b_n$ respectively. This is not true until $$N_k=n_k,\forall k\in\mathbb{N}$$

So, in short, your proof is incomplete. This is exactly what @Lee Chun Min pointed out by using the term 'indices'.


First. This is true, provided that $\limsup a_n$ and $\limsup b_n$ can be added, i.e., the case that one of them is $\infty$ and the other is $-\infty$ should be excluded.

Second. If $\limsup a_n=\infty$ or $\limsup b_n=\infty$, then there is nothing to prove, as the left hand side is equal to $\infty$.

Third. If $\limsup\, (a_n+b_n)=-\infty$, there is also nothing to prove.

So, assuming that both sequences are bounded, we pick a subsequence $$ a_{n_k}+b_{n_k}\to\limsup\, (a_n+b_n). $$ As both $a_{n_k},b_{n_k}$ are bounded, we may pick a common sub-sub-sequence $n_\ell$, such that both $a_{n_\ell},b_{n_\ell}$ converge. Say $$ a_{n_\ell}\to a\le \limsup\, a_n, \qquad b_{n_\ell}\to b\le\limsup\, b_n $$ and hence $$ \limsup\, a_n+\limsup\, b_n \ge a+b=\lim\, (a_{n_\ell}+b_{n_\ell})=\lim\, (a_{n_k}+b_{n_k})=\limsup\, (a_n+b_n) $$


Hint: Assuming $a_n$ and $b_n$ are bounded below !

Now let $a_m + b_m \to \delta =\limsup (a_n + b_n)$, and assume $a_{m_k}$ is a convergent subsequence of $a_m$ in $\Bbb R \cup \{ \infty \}$. WLOG assume $b_{m_k}$ is convergent too (other wise you can extract convergent subsequence from both $a_{m_k}$ and $b_{m_k}$). Now Observe that $a_{m_k} + b_{m_k} \to \delta$. DONE


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.