Help proving that $\lim(s_n+t_n)=+\infty$ 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 $s_n&gt;M$.
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 $s_n + t_n &gt;M$. 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 $s_n&gt;M-t_n$ which implies that $s_n+t_n&gt;M$. 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!
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 $|t_n| &lt; K$ 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 &quot;given some $A$ / given some $\epsilon$ (as appropriate)&quot;. Consider the definition of the statement as a &quot;challenge&quot;: 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 $s_n&gt;M-t_n$). You just have to see that $|t_n|&lt;K$.
You got a definition of a limit to infinity,
check for $s_n+t_n$, that is
$\forall _{M \in \mathbb{R}} \exists _{N \in \mathbb {N}} \forall _{n \geq N} \; s_n + t_n &gt; M \;$ Because...
$\forall _{M \in \mathbb{R}} \exists _{N \in \mathbb {N}} \forall _{n \geq N} \; s_n &gt; M \hspace{30px} \exists _{k,k' \in \mathbb{R}} \forall _{n \in \mathbb {N}} \; k \geq t_n \geq k' \Rightarrow \exists _{K \in \mathbb{R}} \forall _{n \in \mathbb{N}} \; |t_n| &lt; K$
$\Rightarrow \forall _{M \in \mathbb{R}} \exists _{N \in \mathbb{N}}\forall _{n \geq \mathbb{N}} \; s_n &gt; M + K \Rightarrow s_n + t_n &gt; s_n - K &gt; M \quad$ Since $\quad -K &lt; t_n &lt; K$