# If $\{x_n\}\rightarrow \infty$ and $\{y_n\}$ is bounded then $\{x_n+y_n\}\rightarrow \infty$

I'm working on proving the above statement, here's what I have so far.

Proof. Suppose $$|x_n+y_n|$$ is bounded. Then $$\exists \ \varepsilon>0$$, such that $$|x_n+y_n|<\varepsilon$$ for all $$n$$. But $$x_n$$ is unbounded, so $$\forall \ M>0$$, $$\exists \ N\in \mathbb{N}$$ such that for all $$n\geq N$$ implies $$x_n>M$$. Pick $$M>K$$, so that if $$n\geq N$$ $$|y_n|=|(y_n+x_n)-x_n|=|x_n-(x_n+y_n)|\geq |x_n|-|x_n+y_n|>M-K$$

I'm looking for a contradiction, and I don't know where to go from here. I have spent too long on this problem as is, so any help is appreciated.

• $(y_n)$ is bounded, i.e., there exists $d$ such that $|y_n|\leq d$. Put $K=max\{\varepsilon,d\}$ and take $M=3K$ for example. – Jaca Feb 27 '20 at 5:00
• My advice would be to try proving it directly. For all $M > 0$, find some $N$ such that $n \ge N \implies x_n + y_n > M$. You already know you can make $x_n$ as large as you like, and you know that $y_n \ge K$ for some $K$ independent of $n$. – user754697 Feb 27 '20 at 5:01
• Your relation will give you $K\geq |y_n|>M-K=2K$ for all $n\geq N$. – Jaca Feb 27 '20 at 5:01
• If I suppose there exists an $M>0$ such that $-M<y_n<M$, then since $x_n$ diverges to infinity, $x_n>2M$. Am I then able to make the assumption that $x_n+y_n>2M+y_n>2M-M=M$? – drfrankie Feb 27 '20 at 5:08

You can construct a simple proof using comparison test. Suppose $$|y_n| < M$$. Then: $$x_n + y_n > x_n- M \to \infty$$