Question

converge or diverge sum_{2}^{infty} 1/(k ln k)
use integral test

Answer 1

\[ \sum_{k=2}^{\infty} \frac{1}{(k ln k)} \]

Now that looks better.

Answer 2

yes thanks

Answer 3

\[\sum_{k=2}^{\infty}{1\over k\ln k}\]now it looks even better :!
ok let's do it! (looking through notes)

Answer 4

\[ \int_{2}^{\infty}\frac{1}{k \ln k} = ? \]

Answer 5

u-sub ??

Answer 6

idk

Answer 7

yes, u-sub should work

Answer 8

find out that, if that value is less than infinity then it should converge .... says wikipedia
http://en.wikipedia.org/wiki/Convergence_test

Answer 9

if the integral converges so does the series
if the integral diverges, so does the series
that's my understanding

Answer 10

same thing as experimentX said I guess...

Answer 11

@experimentX I'm gonna call you exX from now on, I'm lazy :)

Answer 12

sure .. whatever

Answer 13

i set my u = x
and i got stuck

Answer 14

no let u = ln x

Answer 15

well that ain't the right sub
actually u=x is never the right sub

Answer 16

u=x is redundant
you are just giving x a different name lol

Answer 17

hahaha ... that s funny

Answer 18

okay thats true

Answer 19

well give it a try out ... alert us when you get stuck

Answer 20

so i got \[\lim_{b \rightarrow \infty}\int\limits_{2}^{b} \ln(lnx)\]

Answer 21

am i on the right track???

Answer 22

yup, so does it diverge or converge?

Answer 23

wait a minute... I see something weird...

Answer 24

diverge

Answer 25

because it goes to infinity

Answer 26

do you see the same problem I do exX ?
ln(2)<0
so ln(ln(2)) is undefined

Answer 27

yeah, the other part definitely diverges, but what about evaluation at 2 ?

Answer 28

ln(ln(2)) is just a constant
i thought it shouldn't matter if i subtracted it from infinity

Answer 29

let's ask the wolf first
http://www.wolframalpha.com/input/?i=integrate+from+2+to+infinity+1%2F%28x+lnx%29
wolf says does not converge.

Answer 30

I think it should be rather : ln(ln(inf)) - ln(ln(2))
ln 2 > 1
so lnln2 >1

Answer 31

oh I'm trippin' ^^^

Answer 32

yeah, so divergent

Answer 33

cool

Answer 34

ln2<1
not
ln2<0
like I said befor

Answer 35

Oo ... looks like i didn't see that.

Answer 36

http://www.wolframalpha.com/input/?i=ln%28ln+%282%29%29
it's negative.

Answer 37

but still defined and constant, so it's all good :)

Answer 38

haha ... I forgot 2 < e