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If $\lim s_n=+\infty$ and if $(t_n)$ is a bounded sequence, then $\lim(s_n+t_n)=+\infty$.

My proof:

Suppose $\lim s_n=+\infty$ and $(t_n)$ is bounded.

Since $\lim s_n=+\infty$, by definition we have that for all $M\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies that $s_n>M$.

Since $(t_n)$ is bounded, we have that for all $n\in\mathbb{N}$, $k\leq t_n\leq k'$ for some $k,k'\in\mathbb{R}$.

This is where I get stuck. I'm not sure how to put these two definitions together coherently. It's obvious to me that $t_n$ being bounded means that $s_n + t_n >M$. However, I'm having trouble arriving at this result from the definitions that I've stated.

Maybe since $\lim s_n=+\infty$, I could say that $s_n>M-t_n$ which implies that $s_n+t_n>M$. But I'm not sure how the definition of $t_n$ being bounded plays into this.


Am I on the right track? Any hints/advice would be helpful. Thank you!

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    $\begingroup$ I think you gave the definition of $s_n$ correctly. Maybe since $s_n$ is unbounded, we have that at some point $P$, $s_n>t_n$. Then concluding that $s_n+t_n$>$M$? $\endgroup$ Feb 27 at 23:10
  • $\begingroup$ Try working with a constant with regards to the bounded sequence. More specifically, show that $\lim_{n \rightarrow \infty} [s_n + k]$ diverges. $\endgroup$ Feb 27 at 23:30
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You want to prove that $\lim (s_n + t_n) = \infty$.

Given some $A \in \mathbb{R}$.

We know that there exists $K$ such that $|t_n| < K$ for all $n$.

Since $\lim s_n = \infty$, we can pick $N$ such that $n\ge N\implies s_n \ge A+K$.

Thus, $n\ge N \implies s_n+t_n\ge A \ \ \blacksquare$.

Comment: I used to struggle with these proofs as well. Make sure to think about what exactly you want to prove. I always start these proofs with "given some $A$ / given some $\epsilon$ (as appropriate)". Consider the definition of the statement as a "challenge": if an adversary gives you some $A$, how do you come up with $N$ that beats the challenge?

Further comment on your attempt:

The way you did it almost worked (you got $s_n>M-t_n$). You just have to see that $|t_n|<K$.

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  • $\begingroup$ Ok, cool I think this is exactly what I was looking for. An equivalent way of saying that $t_n$ is bounded is that $|t_n|<K$? $\endgroup$
    – Gteal
    Feb 27 at 23:22
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    $\begingroup$ Well, you should say "there exists K such that...". $\endgroup$ Feb 27 at 23:23
  • $\begingroup$ haha ya I know. I guess I was just blanking on the fact that since its bounded, obviously there exists a $K$ such that $-K<t_n<K$, which is to say that $|t_n|<K$. Thanks for the help! :) $\endgroup$
    – Gteal
    Feb 27 at 23:24
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    $\begingroup$ "$(t_n)$ is bounded" is equivalent to, "There exists K such that $|t_n|<K$ for all $n$". And +1 for this answer: I would have written the proof in the exact same way. $\endgroup$ Feb 27 at 23:25
  • $\begingroup$ In terms of the variables in the original question, it's sufficient to choose $N$ such that whenever $n \ge N$, then $s_n \ge A - k$ (with the thought in the back of one's mind that $k$ might be a "very negative" number, but the boundedness of $t_n$ still ensures that $k$ remains "finitely negative"). $\endgroup$ Feb 28 at 1:14
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We can do a proof by “ contradiction “. Assume $s_n + t_n \to L < \infty $, then $s_n + t_n$ is bounded, say by $M$. Thus $|s_n| = |s_n + t_n - t_n| \le |s_n + t_n| + |t_n| \le M + T$. So $s_n$ is bounded, but it is unbounded since $s_n \to \infty $. Therefore $s_n + t_n \to \infty $.

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    $\begingroup$ And what about the case where $s_n + t_n$ does not approach infinity, but not because it has a finite limit? (For example, to rule out the possibility that $s_n + t_n = (-1)^n n$. Or the possibility that $s_n + t_n = (-1)^n$.) $\endgroup$ Feb 28 at 1:16
  • $\begingroup$ @Daniel Schepler: I thought about the case you brought up but assumed OP’s sequence is either convergent to $\infty$ or to $L < \infty $. $\endgroup$
    – DeepSea
    Feb 28 at 2:23
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You got a definition of a limit to infinity, check for $s_n+t_n$, that is

$\forall _{M \in \mathbb{R}} \exists _{N \in \mathbb {N}} \forall _{n \geq N} \; s_n + t_n > M \;$ Because...

$\forall _{M \in \mathbb{R}} \exists _{N \in \mathbb {N}} \forall _{n \geq N} \; s_n > M \hspace{30px} \exists _{k,k' \in \mathbb{R}} \forall _{n \in \mathbb {N}} \; k \geq t_n \geq k' \Rightarrow \exists _{K \in \mathbb{R}} \forall _{n \in \mathbb{N}} \; |t_n| < K$

$\Rightarrow \forall _{M \in \mathbb{R}} \exists _{N \in \mathbb{N}}\forall _{n \geq \mathbb{N}} \; s_n > M + K \Rightarrow s_n + t_n > s_n - K > M \quad$ Since $\quad -K < t_n < K$

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