if $A_n \longrightarrow \infty $ and $B_n \longrightarrow \infty $ then $(A_n+B_n) \longrightarrow \infty$

Question

if $A_n \longrightarrow \infty $ and $B_n \longrightarrow \infty $

then $(A_n+B_n) \longrightarrow \infty$.

How do you prove it?

Answer 1

Assume not. Than there is a $\text{M}$ such that $\forall n \in \mathbb{N},\, a_n+b_n \leq \text{M} $.

As $A_n \rightarrow \infty$ then $ \exists m, \forall k>m, a_k>\text{M}$.

And the same thing for $b$ too, like:

$B_n \rightarrow \infty$ then $ \exists s, \forall t>s, a_t>\text{M}$.

Then for $\text{max}=\text{MAX}[m,s]$, both $a_{\text{max}}$ and $b_{\text{max}}$ are greater than $\text{M}$. Then $a_{\text{max}}+b_{\text{max}} > \text{M}$, which is a contradiction as we assume that $\exists \text{M}$ such that $\forall n \in \mathbb{N},\, a_n+b_n \leq \text{M} $. Then $A_n+B_n \rightarrow \infty$

Answer 2

Take $$(a_n)=n,~~(b_n)=1-n$$ to see that what will happen.

Answer 3

Assume $a_n\to\infty$ and $b_n$ is bounded from below (which is weaker than $b_n\to\infty$). Let $L=\inf b_n\in\mathbb R$.

Let $M\in\mathbb R$ be given. As $a_n\to\infty$, there exists $N\in \mathbb N$ such that $a_n>M-L$ for all $n>N$. Then for all $n>N$ we have $a_n+b_n>M$, which shows that $a_n+b_n\to\infty$.