Confirmation of a proof that the sum of a bounded and divergent sequence diverges. 
Suppose $y_n\to+\infty$. Suppose also that $\langle x_n\rangle$ is bounded. Show that $x_n+y_n\to+\infty$

With questions that appear this obvious it's always difficult to be sure that what I'm doing is actually correct, and I'm not just being circular in my reasoning. Therefore, I'm looking for confirmation that my attempt at proving the given statement is in fact complete. The attempt is as follows:
Since $\langle x_n\rangle$ is bounded, we have that $\exists\alpha_1,\alpha_2\in\mathbb{R}$ such that $\alpha_1\geq x_n$, $\forall n\in\mathbb{N}$, and $\alpha_2\leq x_n$, $\forall n\in\mathbb{N}$. We may apply transitivity to conclude that $\alpha_2\leq x_n\leq\alpha_1$, $\forall n\in\mathbb{N}$. Therefore, consider
$$
\alpha_2+y_n\leq x_n+y_n\leq\alpha_1+y_n,~\forall n\in\mathbb{N}
$$
Since $y_n\to+\infty$, $\forall m\in\mathbb{R}$, $\exists N$ such that $n>N\implies y_n>m$. We note that we may express $m$ as $p-\alpha_2$, where $p\in\mathbb{R}$. Therefore, we have that $n>N\implies y_n>p-\alpha_2$ or $n>N\implies y_n+\alpha_2>p$. Therefore we have that, by transitivity,
$$
p<\alpha_2+y_n\leq x_n+y_n\leq \alpha_1+y_n
$$
When $n>N$. From this we may conclude that
$$
p<x_n+y_n
$$
When $n>N$. Therefore we have found an $N$ such that $n>N\implies x_n+y_n>p$. We may repeat this process given any $p$, and therefore, we have that $\forall p\in\mathbb{R}$, $\exists N$ such that $n>N\implies x_n+y_n>p$, and thus $x_n+y_n\to+\infty$, as required.
 A: Your proof is fine , just a few comments :


*

*Normally, while starting this kind of a proof , you will start with $p \in \mathbb R$ for which you have to find an $N$. The whole "We may express $m$ as $p - \alpha_2$ where $p \in \mathbb R$" when $p$ has not been mentioned before is a small issue. You come back to it in the end with "we can repeat this for any $p$", but that $p$ should be considered in the starting of the proof logically. How it could have been written is this :



Given $p \in \mathbb R$, let $m = p - \alpha_2$. Then, there exists $N$ so that $n > N \implies y_n > m$ from whence we get $x_n + y_n > p$ for $n > N$. Thus, the given $N$ works for this $p$ and we are done.

See, in words the definition goes "given $p \in \mathbb R$ there is an $N$ ..." so an ideally worded answer starts by taking such a $p$ and then computing the $N$. Your answer has not mentioned the $p$ beforehand : but the logic of the answer is fine.


*

*Note that $x_n$ only needs to be bounded below : you did not use $\alpha_1$, so your proof also shows that the same happens if $x_n$ is only bounded below.

