Help proving that $\lim(s_n+t_n)=+\infty$

Question

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.

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

Answer 1

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 $|t_n|<K$.

Answer 2

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 $.