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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.

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  • $\begingroup$ $(y_n)$ is bounded, i.e., there exists $d$ such that $|y_n|\leq d$. Put $K=max\{\varepsilon,d\}$ and take $M=3K$ for example. $\endgroup$
    – Jaca
    Feb 27, 2020 at 5:00
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    $\begingroup$ 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 $n$. $\endgroup$
    – user754697
    Feb 27, 2020 at 5:01
  • $\begingroup$ Your relation will give you $K\geq |y_n|>M-K=2K$ for all $n\geq N$. $\endgroup$
    – Jaca
    Feb 27, 2020 at 5:01
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    $\begingroup$ 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 $x_n+y_n>2M+y_n>2M-M=M$? $\endgroup$
    – drfrankie
    Feb 27, 2020 at 5:08

1 Answer 1

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

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