real analysis - Absolute Convergence

Prove that if $\vert \frac{a_{n+1}}{a_n} \vert \leq \vert \frac{b_{n+1}}{b_n} \vert$ for $n >>1$, and $\sum b_n$ is absolutely convergent, then $\sum a_n$ is absolutely convergent.

My attempt at a solution:

By the Comparison Theorem, since $\vert \frac{b_{n+1}}{b_n} \vert$ converges by hypothesis, and since $\vert \frac{a_{n+1}}{a_n} \vert \leq \vert \frac{b_{n+1}}{b_n} \vert$, then $\sum a_n$ must also converge.

Questions:

I know this is valid if $b_n$ is not equal to 1. However, how would I proceed to prove this in the case where it is equal to 1? Would I be able to make an estimate using the given information about $b_n$ being absolutely convergent?

Thanks in advance, I greatly appreciate suggestions/feedback/advice.

I am using the textbook Introduction to Analysis by Arthur Mattuck.

Answer

Here is a hint:

You have some $N$ such that For $n \ge N$, $| \frac{a_{n+1}}{a_n} | \le | \frac{b_{n+1}}{b_n} |$.

Then for $n > N$ (notice the strict inequality), and we have $\frac{a_{n}}{a_N} = \frac{a_{N+1}}{a_{N}} \cdots \frac{a_{n}}{a_{n-1}}$, and similarly for the $b_n$.

So $|a_n| \le |a_N| |\frac{a_{N+1}}{a_{N}}| \cdots | \frac{a_{n}}{a_{n-1}} | \le |a_N| |\frac{b_{N+1}}{b_{N}}| \cdots | \frac{b_{n}}{b_{n-1}} | = |a_N| |\frac{b_{n}}{b_{N}}| = | \frac{a_N}{b_N} | |b_n|$.

Addendum: To see why $|\frac{b_{n+1}}{b_n}|$ does not necessarily converge, consider the following:

It is straightforward to see (because of the $\frac{1}{n^2}$ form) that $b_n = \begin{cases} \frac{4}{n^2}, & n \text{ even} \\ \frac{8}{n(n+1)^2}, & n \text{ odd} \end{cases}$ is a convergent series.

If $n$ is odd, we have $\frac{b_{n+1}}{b_n} = \frac{n}{2}$, and hence $\limsup_n \frac{b_{n+1}}{b_n} = \infty$. (And since $\liminf_n \frac{b_{n+1}}{b_n} = 0$, the ratio does not converge.)