# If $x_n \to \infty$ and $y_n$ Cauchy, then $x_n + y_n \to \infty$

Suppose that $(x_n)$ is a sequence of real numbers such that $\displaystyle{\lim_{n \to \infty} x_n = \infty}.$ Suppose also that $(y_n)$ is a Cauchy sequence of real numbers. Show that $$\lim_{n \to \infty} x_n +y_n = \infty.$$

Thoughts: I believe that the result is true because $(y_n)$ is bounded. So you are adding a finite number to infinity. I am not sure if I can cite the vector space property of convergent sequences and that Cauchy sequences are convergent in $\mathbb{R}$ and be done.

And anyways I would like to write a proper $\varepsilon - N$ proof. Well, in this case, it isn't $\varepsilon$, but you get the idea. I have been toying around with inequalities and have even tried a proof by contradiction, but I haven't gotten anywhere with it.

• Since $(y_n)$ is bounded then $y_n\geq l$ for every $n$, for some $l$. Then for every $n$, $x_n+y_n\geq x_n+l$. Since $x_n$ can be made as big as you want (from some point on), why not make it "pretty big+$l$"? – Luiz Cordeiro Sep 8 '14 at 22:13

Hint: Given $M>0$, for sufficiently large $N$, for all all $n>N$ $y_n$ is arbitrarily close a finite number $L$ (it's limit, which exist since the sequence is Cauchy) and $x_n>M$

Here is a proof that doesn't use the fact that $y_n$ converges - only that it is bounded.

Suppose $|x_n + y_n|$ is bounded. Then there exists $K > 0$, such that $|x_n + y_n|\le K$ for all $n$. But $x_n$ is unbounded. So given $M>0$, there exists $N$ such that $\forall n \ge N$, $|x_n| \ge M$. Choose $M \gt K$.

So if $n \ge N$ $$|y_n| = |(y_n + x_n) - x_n| \ge |x_n| -|y_n + x_n|\ge M-K$$

using the triangle inequality. So $y_n$ is unbounded.

Here is a proof which directly uses the fact that $(y_n)$ is Cauchy.

Given $\epsilon > 0$, there is an $N\in \mathbb{N}$ such that $|y_n - y_N| < \epsilon$ for all $n \geq N$.

There is also an $M \in \mathbb{N}$ such that $x_n > M - y_N$ for all $n \geq M$. Therefore, if $n \geq \max\{N,M\}$ then $$|x_n + y_n| > |x_n+y_N| - |y_n-y_N| > M - \epsilon$$ The desired conclusion follows.