If $\limsup\limits_{n\to\infty}a_n=+\infty$ and $\limsup\limits_{n\to\infty}b_n\in\mathbb{R}$, then $\limsup\limits_{n\to\infty}(a_n+b_n)=+\infty$?

Question

There is the following exercise in a book:

$\limsup \limits_{n \to \infty} (a_n + b_n) \leq \limsup \limits_{n \to \infty} a_n + \limsup \limits_{n \to \infty} b_n $

And the author(Kazuo Matsuzaka) writes as follows in his solution without proofs:

If $\limsup \limits_{n \to \infty} a_n = +\infty$ and $\limsup \limits_{n \to \infty} b_n = +\infty$, then $\limsup \limits_{n \to \infty} (a_n + b_n) = +\infty$.

If $\limsup \limits_{n \to \infty} a_n = +\infty$ and $\limsup \limits_{n \to \infty} b_n \in \mathbb{R}$, then $\limsup \limits_{n \to \infty} (a_n + b_n) = +\infty$.

Please tell me how to prove the above two statements if they are true.

Please tell me counter examples if the above two statements are false.

Answer 1

Hint:

You just have to prove that, for all $n\in\mathbf N$, one has: $$\sup_{k\ge n}(a_k+b_k)\le\sup_{k\ge n}a_k+\sup_{k\ge n}b_k $$ then use the basic results on (ordinary) limits.

Answer 2

In this general form, the statements in the solution are incorrect. For the first part, let $a_n = (-1)^n n$ and $b_n = -a_n$. For the second part, let $a_n = \begin{cases}0&n \text{ odd} \\ n, & n \text{ even} \end{cases}$ and $b_n = -a_n$.

Note that in order to prove the original statement $\limsup \limits_{n \to \infty} (a_n + b_n) \le \limsup \limits_{n \to \infty} a_n + \limsup \limits_{n \to \infty} b_n$, these cases are irrelevant. As soon as the right-hand side is $+\infty$, the inequality is automatically correct.

Answer 3

The first is easy to prove:

Since,for all $m \in \mathbb{N}$:

$$ \limsup_{n\to\infty}a_n \leq \sup_{k\geq m} a_k $$ If: $$ \limsup_{n\to\infty}a_n = +\infty $$ Then: $$ \sup_{k\geq m} a_k = +\infty $$ Therefore if: $$ \limsup_{n\to\infty}a_n = +\infty\\ \limsup_{n\to\infty}b_n = +\infty $$ Then, for all $k \in \mathbb{N}$: $$ \sup_{k\geq m}a_n = +\infty\\ \sup_{k\geq my}b_n = +\infty $$ Therefore: $$ \sup_{k\geq m}a_n > 0\\ \sup_{k\geq m}b_n > 0 $$ Therefore: $$ \sup_{k\geq m}a_n + \sup_{k\geq m}b_n > \sup_{k\geq m}a_n $$ By the linearity of $\sup$: $$ \sup_{k\geq m}(a_n + b_n) > \sup_{k\geq m}a_n $$ Taking limits on $k$(allowable since the inequality is valid for all $k \in \mathbb{N}$): $$ \lim_{k\to\infty}\sup_{k\geq m}(a_n + b_n) > \lim_{k\to\infty}\sup_{k\geq m}a_n $$ Thus, by the definition of $\limsup$: $$ \limsup_{n\to\infty}(a_n + b_n) > \limsup_{n\to\infty}a_n $$ Since: $$ \limsup_{n\to\infty}a_n = +\infty $$ Therefore: $$ \limsup_{n\to\infty}(a_n + b_n) = +\infty $$

I'll have to think about the second statement.