I'm currently working through problems given in Mathematical Methods for Physicists by Arken et al. I'm having a bit of trouble resolving a problem that comes up in Chapter 1. It is as follows:
If $\lim_{n\rightarrow\infty} \frac{b_n}{a_n} = K$, a constant with $0 < K < \infty$, show that $\sum_{n}b_n$ converges or diverges with $\sum a_n$.
Provided Hint: If $\sum a_n$ converges, rescale $b_n$ to $b_n' = \frac{b_n}{2K}$. If $\sum a_n$ diverges, rescale to $b_n'' = \frac{2b_n}{K}$.
Plugging everything in, I'm left with the following expression: $$ \lim_{n\rightarrow\infty} \frac{b_n}{a_n} = K \longrightarrow \lim_{n\rightarrow\infty} \frac{b_n'}{a_n} = \frac{1}{2} \\ $$
From here, I would usually proceed to split the limit and rearrange like so: $$ \frac{\lim_{n\rightarrow\infty}b_n'}{\lim_{n\rightarrow\infty}a_n} = \frac{1}{2} \\ \lim_{n\rightarrow\infty}b_n' = \frac{1}{2} \lim_{n\rightarrow\infty} a_n $$
In this case, I realise I can't do that because of the fact that under the assumption that $\sum a_n $ converges, $\lim_{n\rightarrow\infty} a_n$ must be equal to zero by the limit test.
Where have I messed up on my reasoning?
Thanks!
Your proposition is incorrect as stated.
If $\sum a_n$ and $\sum b_n$ are series with positive terms then the result
Given $$ \lim_{n\rightarrow\infty} \frac{b_n}{a_n} = K\text{ with }0 < K < \infty $$ implies that $$\sum b_n \text{ converges or diverges with } \sum a_n $$ is well known as the Limit Comparison Test (Wikipedia has a proof).
If you leave out the assumption that $a_n,\ b_n > 0$ then there are counter-examples to the conclusion. From Nguyen S. Hoang's A Limit Comparison Test for General Series:
If $a_n = \frac{(-1)^n}{\sqrt{n}}$ and $b_n = \frac{(-1)^n}{\sqrt{n}} + \frac{1}{n}$ then $\frac{b_n}{a_n} \rightarrow 1$ and $\sum a_n$ converges as an alternating series, but $\sum b_n$ diverges.