Prove that $\limsup_{n \to \infty} \left(a_n + b_n \right) \leq \limsup_{n \to \infty} a_n +\limsup_{n \to \infty} b_n$

Question

I want to prove that for two sequences, say $a_k$ and $b_k:$

$$ \limsup_{n \to \infty} \left(a_n + b_n \right) \leq \limsup_{n \to \infty} a_n +\limsup_{n \to \infty} b_n$$

If we let $M_n =\sup \{ a_k : k \geq n \}$ and $N_n = \sup \{ b_k : k\geq n \} $, then obviously

$$a_k +b_k \leq M_n+ N_n \Leftrightarrow \sup \{a_k+b_k: k \geq n \} \leq M_n + N_n$$

After that, the order limit theorem should suffice, right? Thanks.

Yes, that's correct +1. I made a few cosmetic changes, I hope it's ok. — Julien, Dec 26 '13 at 17:27
About your rollback. To denote the limsup of a sequence $x_n$, you are supposed to write $\limsup_{n\rightarrow +\infty} x_n$ and not $\limsup_{n\rightarrow +\infty} x_k$. Indices should be the same, otherwise it would mean that the sequence is constant. — Julien, Dec 26 '13 at 17:37

score 2 · Accepted Answer · answered Dec 26 '13 at 17:34

2

You have defined $M_n$ and $N_n$ and you know that $$ \lim_{n\to\infty}M_n=\limsup_{n\to\infty}a_n\quad\text{and}\quad \lim_{n\to\infty}N_n=\limsup_{n\to\infty}b_n. $$ Next let $$ P_n=\sup\{a_k+b_k: k\ge n\}, \quad\text{and likewise}\quad \lim_{n\to\infty}P_n=\limsup_{n\to\infty}(a_n+b_n). $$ As you have noticed $$ P_n\le M_n+N_n. $$ Taking the limit in the above you are done.

answered Dec 26 '13 at 17:34

Yiorgos S. Smyrlis

83,933

Why is $P_n\le M_n+N_n$ – Thatπthere Mar 02 '24 at 00:42
@Thatπthere You could have a look at: Supremum of sum of two sequences: $\sup (x_n+y_n) \le \sup x_n + \sup y_n$ – Martin Sleziak Mar 02 '24 at 11:12

Prove that $\limsup_{n \to \infty} \left(a_n + b_n \right) \leq \limsup_{n \to \infty} a_n +\limsup_{n \to \infty} b_n$

1 Answers1