Proving that if $\lim\limits_{n\to\infty}a_n=+\infty$ and $\{b_n\}$ is a bounded sequence, then $\lim\limits_{n\to\infty}(a_n+b_n)=+\infty$

Question

I am trying to prove the following problem

If $\lim\limits_{n\to\infty}a_n=+\infty$ and $\{b_n\}$ is a bounded sequence, then $\lim\limits_{n\to\infty}(a_n+b_n)=+\infty$

I have these definitions as tools;

Definition 1. The sequence $\{a_n\}\to \infty$ if $\forall \;M\in R,\;\;\exists\;n_0=n_0(M)\in N$ $\ni$ $$n\geq n_{0}\;\implies\;a_n>M.$$

Definition 2. The sequence $\{b_n\}$ is said to be bounded if there exists $M>0,$ $\ni$ $$|b_n|\leq M \;\forall \;n\geq 1.$$

Can anyone help me out?

Answer 1

We know that {b_n} is bounded therefore there exists a $B>0$ such that $-B< b_n< B $ for all $n\in \mathbb N$

Let $M>0$ be an arbitrary real number.

Since $\{a_n\}$ diverges to $\infty$, there exists some positive integer $N$, such that $$ n\ge N \implies a_n >M+B$$

Now if $ n\ge N $ we have $$a_n >M+B \implies a_n+b_n >M+B -B=M $$

Thus $ \{a_n + b_n \}$ diverges to $\infty $

Answer 2

Let $B$ be the bound for $(b_n)$. For any $M\gt0$ choose $n_0\in\mathbb N$ such that $n\ge n_0 \implies a_n\gt M+B$. Then $a_n+b_n\gt M$...

Answer 3

HINT: Rewriting part of your definitions:

1. $\forall N\exists n_0\ \forall n\geq n_0\ a_n>N$.
2. $\forall n |b_n|\leq M$.

It means that $a_n+b_n\geq N-M$ for large $n$.