I am asking how to do the Limit comparison test for this: $$\lim {An\over Bn} =L$$ You choose $B_n$ yourselve, but how do you choose it?
Example:
$$A_n = \frac{3n^2 + 5n + 1}{\sqrt{(n^5 + 5 )}}$$
$$B_n = {3\over \sqrt{n}}$$
$$\large \lim {\frac{3n^2 + 5n + 1}{\sqrt{(n^5 + 5 )}}\over {3\over \sqrt{n}}} = 1$$
It can be concluded that $\lim A_n$ diverges.
Why? How to get $B_n$
and how to do the limit comparison test for this?
An example of another question: $$An={(2n+200)\over (e^\frac n3)-20}$$ An converges or diverges?
$\endgroup$ 43 Answers
$\begingroup$Look at $A_n$. On the numerator the dominating term in $3n^3$, and on the denominator its $\sqrt{n^5}=n^2 \sqrt{n}$. Dividing the two gives $$\frac{3n^3}{n^2 \sqrt{n}}=3 \sqrt{n}. $$
$\endgroup$ 3 $\begingroup$In such excercises, you shall take Bn to be a series that is "simpler" than what you have, but of same "order". To do so, you shouldl, roughly, spot the strongest parts of the numerator and the denominator, and divide them.
$\endgroup$ 3 $\begingroup$For $n$ sufficiently large, you can expand and get $$A_n = \frac{3n^3 + 5n + 1}{\sqrt{n^5 + 5 }}= \sqrt n \frac{3n^3 + 5n + 1}{\sqrt{n^6 + 5n }}\simeq \sqrt n \frac{3n^3}{\sqrt{n^6 }}=3 \sqrt n $$
More sophisticated would be a Taylor expansion $$A_n=3 \sqrt{n}+5 \left(\frac{1}{n}\right)^{3/2}+\left(\frac{1}{n}\right)^{5/2}+O\left(\left(\frac{1 }{n}\right)^{9/2}\right)$$
There is clearly a typo somewhere.
Added later fater the typo was fixed
If $$A_n = \frac{3n^2 + 5n + 1}{\sqrt{n^5 + 5 }}$$ it is obvious to ntoice that, for large values of $n$, $A_n \approx \frac{3n^2}{n^{5/2}}=\frac{3}{\sqrt n}$. THe expansion would be $$A_n=3 \sqrt{\frac{1}{n}}+5 \left(\frac{1}{n}\right)^{3/2}+\left(\frac{1}{n}\right)^{5/2}+O\left(\left(\frac{1 }{n}\right)^{7/2}\right)$$
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