Limit Comparison Test with upside-down division

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

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?

• It doesn't matter much whether you take the limit of $a_n/b_n$ or of $b_n/a_n$. In this example, I wouldn't choose $b_n=1/n^2$ though. Commented Apr 6, 2019 at 17:19
• It's the first series that came to mind that converges, but it was just an example though. The more important thing was to be able to assume I can take this test either way :) Commented Apr 6, 2019 at 17:38
• Im kind of stuck, do you have any suggestion which series is a good pick for the test? Commented Apr 7, 2019 at 12:58

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.

• thank you for the informative proof of the first part! :) the series you offered for the test is quite good with the justification the there is an index n from which the series is positive. Ultimately I get a $\lim_{n \to \infty} \frac{a_{n}}{b_{n}}=1$ but I figured out another way to solve it using the fact that both parts of the fraction act like a polynomial with rank 'r' and if $r_{2}-r_{1}> 1$ the series converges Commented Apr 8, 2019 at 5:18