Limit of sum of unbounded and bounded sequence Suppose $a_n \rightarrow +-\infty$ and $(b_n)$ is bounded. Show that $a_n+b_n \rightarrow +-\infty$. I tried this:
$|a_n|\rightarrow +-\infty$, so $|a_n+-\infty|<\epsilon$. It is also true that $|b_n|<M$, because $(b_n)$ was bounded. Now, if you add those together, you will get:
$|a_n+-\infty|+|b_n|<\epsilon+M$ and therefor: $|a_n+b_n+-\infty|\le|a_n+-\infty|+|b_n|<\epsilon+M$. But, since M is fixed (say M was $2$), you can't get closer than $2$ to the limit (because $\epsilon$ must be negative in that case). Am I missing something?
Regards,
Kevin
 A: I wouldn't advise you to add/subtract infinity until you'll have enough experience in this. 
The strict proof is like this: 


*

*suppose that $a_n\to+\infty$, so for any $E>0$ (here we are especially interested in large values of $E$) there exists $N(E)$ such that $a_n>E$ for all $n\geq N$. 

*As you have written, there is a constant $M$ such that $|b_n|<M$ for all $n\geq0$, i.e.
$$
-M<b_n<M.
$$
We will use the left-hand side of the inequality.

*To prove that $a_n+b_n\to+\infty$ we should show that for any $E'$ there is $N(E')$ such that for all $n\geq N(E')$ holds $a_n+b_n>E'$. 

*We can clearly do it: pick up any $E'$, then $a_n+b_n>a_n-M$ (see 2.), hence to make $a_n+b_n>E'$ we just need to make $a_n>E'+M$ for any $E'$ - and that will be sufficient (do you agree here?)

*Based on 1., we just take $N(E'+M)$ so $a_n>E'+M$ for all $n\geq N(E'+M)$, hence
$$
a_n+b_n>E'
$$
for all $n\geq N(E'+M)$ and hence $a_n+b_n\to +\infty$.
Could you please follow the same steps to prove the case when $a_n\to -\infty$?
