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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.

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    $\begingroup$ 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 '17 at 8:45
  • $\begingroup$ I am sorry but I didn't quite understand. "The problem is about indices" $\endgroup$ Jun 17 '17 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 '17 at 8:54
  • $\begingroup$ You must add more arguments, like $a_n ,b_n$ are non-negative sequences $\endgroup$
    – Red shoes
    Jun 17 '17 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 '17 at 9:22
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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 $$ \limsup\, a_n+\limsup\, b_n \ge a+b=\lim\, (a_{n_\ell}+b_{n_\ell})=\lim\, (a_{n_k}+b_{n_k})=\limsup\, (a_n+b_n) $$

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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'.

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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

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