1
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Commented Apr 6, 2019 at 17:19
  • $\begingroup$ 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 :) $\endgroup$
    – Avi P
    Commented Apr 6, 2019 at 17:38
  • $\begingroup$ Im kind of stuck, do you have any suggestion which series is a good pick for the test? $\endgroup$
    – Avi P
    Commented Apr 7, 2019 at 12:58

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – Avi P
    Commented Apr 8, 2019 at 5:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .