$a_n$ and $b_n$ are assumed to bounded sequences. The definition I'm strictly adhering to is

$\varliminf a_n = \inf\{\text{subsequential limits of } a_ n \} $

$\varlimsup a_n = \sup\{\text{subsequential limits of } a_ n \} $

Below is an attempt I made; would highly appreciate any feedback on it. I have both seen and done other proofs of this statement, using alternative definitions of limit superior/inferior, so I wanted to write up a new one on my own.

Let $L_2$ be any subsequential limit of the sequence $b_n$. Then some $b_{n_k} \to L_2$.

Study the sequence $a_{n_k}$; this is bounded and thus has a convergent subsequence $a_{l_k}$. Denote the limit $L_1$. We know $\varliminf a_n \leq L_1$.

Looking at exactly the $l_k$-indices of $b_{n_k}$, we know $b_{l_k}$ is a subsequence of $b_{n_k}$ and hence also converges to $L_2$.

But then $a_{l_k} + b_{l_k}$ must converge to $L_1 + L_2$. But then of course $L_1 + L_2 \leq \varlimsup (a_n + b_n)$.

The statement we have is then: $\varliminf a_n + L_2 \leq L_1 + L_2 \leq \varlimsup (a_n +b_n) $, that is

$L_2 \leq \varlimsup (a_n +b_n) - \varliminf a_n$

So we have found an upper bound to the set of subsequential limits of $b_n$, which finally leaves us with:

$\varlimsup b_n \leq \varlimsup (a_n + b_n) - \varliminf a_n$. We are done $\blacksquare$.


1 Answer 1


Your proof is fine. I only would use a different notation for the “iterated subsequence” (that is what caused my initial confusion): If $(b_{n_k})$ is a convergent subsequence of $(b_n)$ with limit $L_1$ then we can choose a convergent subsequence $(a_{n_{k_j}})$ of $(a_{n_k})$. Then $$ L_2 = \lim_{j \to \infty} a_{n_{k_j}} \ge \varliminf a_n $$ and $$ L_1 + L_2 = \lim_{j \to \infty} (a_{n_{k_j}} + b_{n_{k_j}}) \le \varlimsup (a_n + b_n) $$ so that $$ L_1 = (L_1 + L_2) - L_2 \le \varlimsup (a_n + b_n) - \varliminf a_n $$ This holds for all convergent subsequences of $(b_n)$, which implies $$ \varlimsup b_n \le \varlimsup (a_n + b_n) - \varliminf a_n \, . $$

  • $\begingroup$ Thank you so much, cheers! $\endgroup$
    – kapython
    Feb 4, 2021 at 20:24

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