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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.

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  • $\begingroup$ "...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 $b_n = 1/n^2$.. $\endgroup$ – Adam Rubinson Dec 4 '20 at 1:29
  • $\begingroup$ I'm not sure what the Algebraic Limit Theorem is. Please can you state clearly what it is (preferably, write it in your question). $\endgroup$ – Adam Rubinson Dec 4 '20 at 1:35
  • $\begingroup$ 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. $\endgroup$ – Oreomair Dec 4 '20 at 1:35
  • $\begingroup$ @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? $\endgroup$ – ml4592 Dec 4 '20 at 1:37
  • $\begingroup$ @Oreomair you're right, thank you for the clarification! $\endgroup$ – ml4592 Dec 4 '20 at 1:42
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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.

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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.:

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

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  • $\begingroup$ 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 $|\frac{a_n}{b_n} -6| < \epsilon$ $\endgroup$ – ml4592 Dec 4 '20 at 1:51
  • $\begingroup$ 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. $\endgroup$ – Adam Rubinson Dec 4 '20 at 2:05
  • $\begingroup$ 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. $\endgroup$ – ml4592 Dec 4 '20 at 2:08

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