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Suppose $(s_n)$ is a convergent sequence and $(t_n)$ diverges to $\infty$. Prove that lim $s_n+t_n = \infty$.

Proof (can someone verify it?): Pick $N_1$ such that $\forall n > N_1$, $|s_n-s|<1$. Then, $\forall n > N_1$, $s_n > s-1$.

Now, let $M > 0$. Pick $N_2$ such that $\forall n > N_2$, $t_n > M -s+1$.

Let $N = $ max$\{N_1, N_2\}$. Then $\forall n > N$,

$t_n+s-1>M$. So, $t_n+s_n > M$

Therefore, lim $s_n + t_n = \infty$

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  • $\begingroup$ Yes, well done. $\endgroup$ – user61527 May 17 '14 at 4:05
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    $\begingroup$ Excellent. You will notice that your proof applies more generally for $(s_n)$ bounded from below $\endgroup$ – Hagen von Eitzen May 17 '14 at 4:07
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Your proof is correct!

As Hagen von Eitzen notes in the comments, the core of the proof is the fact that $s_n$ is eventually bounded below. And "eventually bounded below" and "bounded below" are the same thing.

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You are incorrect. The definition of a diverging series is $\forall T>0$, $\forall N\in\mathbb{N}$, $\exists n>N$, s.t. $a_n>T$

This is a "frequently" definition, in contrast to the "eventually" definition of a converging series. However, your idea is very close

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