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I wish to prove that if $(x_n)$ and $(y_n)$ diverge to positive infinity, then $x_n + y_n$ diverges to positive infinity as well.

I started off with definition of divergence. That is: For any real number $M_1$, there exists a real number $N_1$, $n>N_1$ such that $x_n>M_1$. For any real number $M_2$, there exists a real number $N_2, n>N_2$ such that $y_n>M_2$ Now, consider a sequence $(z_n)$ such that $z_n = x_n + y_n$. To prove that it diverges to positive infinity I need to find such $N$, $n>N$ that for any real number $M$, $z_n>M$.

Since $x_n$ and $y_n$ diverge, then $x_n>M_1$ as well as $x_n>M_2$. Similarly, $y_n>M_2$ as well as $y_n>M_1$.

After that, I am stuck as I don't see the way how I can find $N$ for $z_n$.

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  • $\begingroup$ First of all, note the following: in any of the three cases, for any real number $M$ you want to find an $N$ such that $\dots>M$ for all $n>N$. It's not so clear from what you're writing. $\endgroup$ – Ben Nov 3 '16 at 23:41
  • $\begingroup$ Your wording of the definition is a bit imprecise. Anyway, you can just choose the max of the two M's. $\endgroup$ – SBareS Nov 3 '16 at 23:42
  • $\begingroup$ It would help your argument if you had fewer $M$'s. You don't get anything out of potentially having $M_{1}\neq M_{2}$. $\endgroup$ – JessicaK Nov 3 '16 at 23:45
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Since $\lim_{n\rightarrow \infty} y_n = \infty$, there exists $N_1$ such that $n \geq N_1 \Rightarrow y_n >1$.

Let now $M > 0$. Since $\lim_{n \rightarrow \infty} x_n = \infty$ there exists $N_2$ such that $n \geq N_2 \Rightarrow x_n > M$.

Now let $N = max \{N_1, N_2\}$. Then $n \geq N \Rightarrow x_n > M$ and $y_n > 1$, so $x_n + y_n > M+1 > M$.

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