Question

does this converge or diverge? full solution please

Answer 1

\[a_n = \frac{[\ln(n)]^2}{n}\]

Answer 2

Use the comparison test

Answer 3

The full solution is in your head. What is your plan? Did you try the Ratio Test? There are relatively simple comparisons.

Answer 4

It's another indeterminate form, dude.

Answer 5

I don't know the ratio test and I have trouble with the ^2 sign. What do I do with it?

Answer 6

Chain rule.

Answer 7

\[
([\ln(n)]^2 )' = \left( \frac {d}{du}u^2 \right)\Bigg|^{u=\ln(n)} \left(\frac{d}{dn}\ln(n)\right)
\]

Answer 8

You do not need any of that fancy machinery.

\(\dfrac{(ln(n))^{2}}{n} > \dfrac{1}{n}\) for n > 2. Done.

Answer 9

\[
= 2u \Bigg|^{u=\ln(n)}\frac 1n = 2\ln(n)\cdot \frac{1}{n}
\]

Answer 10

The bottom just becomes \(1\). We have another indeterminate form, so we just do it again.

Answer 11

Let me sit on this for a while

Answer 12

Chain rule bothers you?

Answer 13

Try product rule: \[
\ln(n)\cdot \ln(n) = \frac{1}{n}\ln(n) + \frac{1}{n}\ln(n) = \frac{2}{n}\ln(n)
\]

Answer 14

@mebs I strongly suggest you take heed of what tkhunny said, use the comparison test

Answer 15

@Silent_sorrow, I understand that using the comparison test and other methods like the squeeze TH but, up to this part of the chapter we don't "know that stuff".

Answer 16

@wio Im trying to justify why I should differentiate the top and bottom. the top approaches infinity and bottom approaches infinity so that's why we use L'hospital?

Answer 17

Yes, since \(\infty/\infty\) is an indeterminate form.

Answer 18

No objection to l'hospital. I've always thought that l'hospital was more of a last resort than a primary method. Anyway, as I have said many times, unique results don't care how you find them. There are reasons why mathematicians over the years have invented various methods. It would behoove all to learn all and keep an open mind.