# 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.

• "...we know that division by $lim(b_n)$ is valid, so $lim(b_n) \ne 0$". No. Surely $a_n$ and $b_n$ can both tend to $0$, e.g. $a_n = 6/n^2$ and $b_n = 1/n^2$.. Commented Dec 4, 2020 at 1:29
• I'm not sure what the Algebraic Limit Theorem is. Please can you state clearly what it is (preferably, write it in your question). Commented Dec 4, 2020 at 1:35
• You cannot quite use the algebraic limit theorem that way, it supposes that $\lim(b_n)$ exists, which is what you're trying to prove. Commented Dec 4, 2020 at 1:35
• @AdamRubinson thank you for the clarification! That helps with the first have of my confusion, however I'm still confused about the second part. By the Algebraic Limit Thm we have $lim \frac{a_n}{b_n} = 6 \rightarrow \frac{lim(a_n)}{lim(b_n)}$. But by the theorem mentioned, if we know $\sum a_n$ converges then $lim (a_n)$ must equal 0, correct? Commented Dec 4, 2020 at 1:37
• @Oreomair you're right, thank you for the clarification! Commented Dec 4, 2020 at 1:42

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.

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?



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.:

$$\frac{a_n}{7} < b_n < \frac{a_n}{5} \quad \forall n \geq N.$$

• Can you describe how you got there? Using the formal definition of a limit, I understand that $\forall \ \epsilon > 0 \ \exists \ N \in \mathbb{N}$ such that $|\frac{a_n}{b_n} -6| < \epsilon$ Commented Dec 4, 2020 at 1:51
• It's kind of "obviously true" if you're familiar with limits. However, I will modify my answer shortly to give a formal proof of my statement. Commented Dec 4, 2020 at 2:05
• Thank you! My professor is quite the stickler for proofs, so I like to make sure I fully understand the exercises without taking any shortcuts, however obvious they may be. Commented Dec 4, 2020 at 2:08