Prove $\sum b_n$ converges if $\sum a_n$ converges and $\lim \frac{a_n}{b_n}=6$ I recently received this problem on a real analysis exam and I'm still having a hard time understanding it. 

"Let $(a_n)$ and $(b_n)$ be sequences of $\mathbb{R}^+$. Suppose $\lim \frac{a_n}{b_n}=6.$ Prove that if the infinite series $\sum a_n$ converges, then $\sum b_n$ converges."
Here's what I understand so far: we can use the Algebraic Limit Theorem to simplify and get $\lim (a_n) = 6*\lim(b_n)$. Also, since we know $\lim \frac{a_n}{b_n}=6$, we know that division by $\lim(b_n)$ is valid, so $\lim(b_n) \ne 0$.
What's confusing me is that there is a theorem that says if an infinite series $\sum s_n$ converges, then $\lim(s_n) =0$. If we attempt the problem and suppose that $\sum a_n$ converges, wouldn't that mean that $\lim(a_n) =0$, so $\lim \frac{a_n}{b_n}=6$ can't occur? What am I missing?
Edit: I understand now that the Algebraic Limit Thm may not come into play with this problem. However, I'm still confused how to use the given information to prove that $\sum b_n$ converges.
 A: Hint: Let $\epsilon>0$. Since $\frac{a_n}{b_n} \to 6,$ there exists $N$ such that $\frac{a_n}{b_n}>6-\epsilon$ for all $n\geq N$. Show that $\sum b_n$ is bounded above. It is obviously monotone.
A: To prove it, I would do the following:
$\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right) = 6 \implies \exists N \in \mathbb{N}$ such that $\frac{a_n}{7} < b_n < \frac{a_n}{5} \quad \forall n \geq N.$
Can you finish this now?
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Formal proof of the above statement, as requested by OP:
By definition of the limit of a sequence, given $\varepsilon>0, \ \exists N \in \mathbb{N}$ such that $\left|\frac{a_n}{b_n} - 6\right| < \varepsilon \quad \forall \ n \geq N.$
Let $\varepsilon = \frac12.$
Then $ \exists N$ such that:
$\left|\frac{a_n \ - \ 6b_n}{b_n}\right| < \frac12 \quad \forall n \geq N,\ $ which implies that
$ \left|a_n-6b_n\right| < \frac12 b_n \quad \forall \ n \geq N \quad (*),$
where $|b_n| = b_n$ is justified because $b_n \in \mathbb{R}^+$ was assumed in the question.
If $a_n > 6b_n,\ $ then $\left|a_n-6b_n\right| = a_n - 6b_n$, so $(*) \implies a_n < 6 \frac12 b_n\ < 7b_n \quad \forall \ n \geq N.$
Else if $a_n \leq 6b_n\ ,$ then $\left|a_n-6b_n\right| = -a_n + 6b_n$, so $(*) \implies a_n > 5 \frac12b_n\ > 5b_n \quad \forall \ n \geq N.$
So $5b_n < a_n < 7b_n \quad \forall \ n \geq N$, i.e.:
$\frac{a_n}{7} < b_n < \frac{a_n}{5} \quad \forall n \geq N.$
