I have a series $\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{\sqrt[2019]{n+2020}}{n^2-2020}$ and I'm looking for a series that converges in order to use the Limit Comparison Test, such as: $\sum_{n=1}^{\infty}b_{n} = \frac{1}{n^2}$ , would that be ok to use the test while dividing upside-down?
Usually the explanations about the test require that I devide the series$\sum_{n=1}^{\infty}a_{n}$ that I have by the one I call $\sum_{n=1}^{\infty}b_{n}$, and finding if $\lim_{n \to \infty} \frac{a_{n}}{b_{n}}\rightarrow k$ , whereas $0<k<\infty.$
Assuming THERE IS suck k, then the $\lim_{n \to \infty} \frac{b_{n}}{a_{n}}$ should be $\frac{1}{k}$ by arithmetics, and in case k is not $0$ nor $\infty$, can I rely on arithmetics and use this limit as well and make conclusions based on the comparison test, can't I?
Suppose I need to use the Limit Comparison Test to decide if $\sum_{n=1}^{\infty}a_{n}$ coverges , am I restricted only to the form of $\lim_{n \to \infty} \frac{a_{n}}{b_{n}}$ , or the assumptions that I made were eligible and I can use the upside-down version as well?
Firstly, it does not matter if you divide them in the other direction.
Proof: Suppose $\lim_{n \to \infty} \frac{a_{n}}{b_{n}}=k$, where $0 \lt k \lt \infty$. Note that since the sequence $A=(a_n), B=(b_n)$ are greter than $0 $ (strictly).
Consider the sequence $C=(\dfrac{a_n}{b_n} \cdot \dfrac{b_n}{a_n})=(1)$, which converges to $1$.
By limit theorem, $\lim_{n \to \infty} \frac{b_{n}}{a_{n}}$ exists and equal $(lim (1))/\lim_{n \to \infty} \frac{a_{n}}{b_{n}}=1/k$. Which is again strictly greater than $0$, and finite.
Hence it happens that $\lim_{n \to \infty} \frac{a_{n}}{b_{n}}=k$, where $0 \lt k \lt \infty$ iff $ \lim_{n \to \infty} \frac{b_{n}}{a_{n}}=1/k$ where $0 \lt 1/k \lt \infty$.
Secondly, for a good series to solve the problem, consider $ \sum_{n=1}^{\infty}b_{n} = \frac{ ^{2019}\sqrt{x}}{n^2}$.
Note that since $a_n$ are not positive in the first few terms ($\lfloor \sqrt(2200) \rfloor$ terms), truncate it such that all terms are positive, hence condition of limit comparison test is fulfilled.