Question

Determine if the sequence converges: [ln(n)]^2 /n

Answer 1

since \((\ln n)^2>1\) then,
\[\frac{(\ln n)^2}{n}>\frac{1}{n}\]
thus,
\[\sum_{n=0}^N \frac{(\ln n)^2}{n}>\sum_{n=0}^N \frac{1}{n}\rightarrow\infty\] as \(N\rightarrow\infty\) so the series diverges

Answer 2

Thanks for the explanation of this it is interesting :)

Answer 3

@ljensen This is a SEQUENCE. What you solved for was a series.
Even though I know the answer, i dont understand how they got "0"

Answer 4

limit n->inf (ln n)^2/n = 0, so the limit converges,
use L'hospital rule to calculate limit.

Answer 5

\[\lim_{n \rightarrow \infty} \frac{[\ln (n)]^{2}}{n} =\frac{2[\ln(n)]}{n} = \frac{2/n}{1}= 2/n=0\]

Answer 6

yeah that's it.

Answer 7

although you are quite missing to put limit ... until the second last step.

Answer 8

I know; It was too much typing, lol. I know to use L'hospital rule and chain rule or whatever. But im confued on the second step.

The 2[ln(n)]/n is supposed to be 2[ln(n)]/n/1 and i dont understand why

Answer 9

that's all right ... since n from ln n is gong down, and the other n at bottom becomes 1, so you have 2[ln(n)]/n

Answer 10

Thats the thing, lol. Where's that other "n" come from?!

Answer 11

\[ \frac{d(\ln n)^2}{dn} = 2 \ln n\ [d(\ln n)/dn] \] <--- comes from chain rule

Answer 12

0___0
Im an idiot, smh. Thank you.

Answer 13

you're not idiot ... sometimes everyone get's confused.
and you are welcome.