Question

\[\sum_{}^{}\frac{ \ln k }{ k^2 }\]

Does this converge, or diverge?

Answer 1

Do I have to use the integral test here, or is there another way?

Answer 2

ln(x) grows slower than any positive power of x.
So the top is growing much much much slower than the bottom.
So we can at least say this,\[\large\rm \lim_{k\to\infty}\frac{\ln k}{k^2}=0\]

Answer 3

I'm not sure about what test and all that business though,
thinking :d

Answer 4

yea for sure.

Answer 5

Thats how one of my classmates did the problem

Answer 6

I know you want to bump ahead of me. Do it. I dare you.

Answer 7

HAA GOTTEEEM

Answer 8

any ideas?

Answer 9

I wonder if you can just do a comparison test... since \[\large \ln k < k^b\]where b<1, then\[\large \frac{ \ln k }{ k^2 } < \frac{ k^b }{ k^2 }=\frac{ 1 }{ k^{2-b} }\]and 2-b > 1, and isn't it essentially known that \[\large \sum_{n=1}^{\infty}\frac{ 1 }{ k^a }\] converges if a>1...

Probably not "technically correct" enough, but it makes perfect sense to me.

Answer 10

hm, makes sense to me too agent

Answer 11

From looking at some graphs, \[\large \ln k < k^{0.4}\]but\[\large \ln k < k^{0.3}\]is not true.

Now I wonder if there's an easy way to find for what value of w\[\large \ln k < k^{w}\]ceases to always be true. I kinda doubt it, without using that damn Wronskian.

Answer 12

agh

Answer 13

\(\large\rm ln k < k^{0.3}\)
This is still true ^
you just have to start from a larger k value.

Answer 14

I mean for all values of k.

It turns out it's 1/e.

Answer 15

Oh I see

Answer 16

http://www.wolframalpha.com/input/?i=ln(x)%3Cx%5E(m)


1/e ≈ 0.368 so it makes sense, given it's a true statement for m=0.4 and not always true for m=0.3

Answer 17

So we know\[\Large \frac{ \ln k }{ k^2 } \le \frac{ 1 }{ k^{2-\frac{ 1 }{ e }} } \]
And as above that should converge, and does... and hey look, by the comparison test:
http://www.wolframalpha.com/input/?i=sum+from+1+to+infinity+of+%5Cfrac%7B+1+%7D%7B+k%5E%7B2-%5Cfrac%7B+1+%7D%7B+e+%7D%7D+%7D

And hey look, comparison test!
http://www.wolframalpha.com/input/?i=sum+from+1+to+infinity+of+ln+(k)%2Fk%5E2

Answer 18

Hm interesting. I didnt think it would be so involved, but it makes sense. Thank you @agent0smith and @zepdrix for the explanations. Trying my best to cram before my final this saturday x_x

Answer 19

Oooo that's soon :O Exciting!

Answer 20

@legomyego180 I don't think it needs to be this involved... just, like @zepdrix just said (or didn't really say, but I misinterpreted...), it got exciting and interesting so I pursued it.


http://www.wolframalpha.com/input/?i=ln(x)%3Cx%5E(1%2Fe)

And it makes sense from looking at it, that the only equality point is e^e!

Answer 21

you didn't put any limits in there, for starters! so if it's \(k \in [4, 9]\) you're probably safe whatever the series!!

but i think you were right to start with with the integral test. it's an easy integration, so you just needed to integrate to show that it converges.

Answer 22

Ah!! your classmates already did it. great.

Answer 23

the sum is unbounded in the problem, so Im assuming \[k \in [1,\infty)\]

btw how do you guys type your latex formulas so that they dont create a break in the sentence?

Answer 24

@IrishBoy123 That's boring. You're boring everybody. Quit boring everyone! https://www.youtube.com/watch?v=B35vsgBzMh4

Answer 25

So taking the improper integral would be a little more straight forward?

@agent0smith boring = easy and I like easy lol

Answer 26

I still think the comparison test is easy too, though. You don't need anything beyond my first post (i just became intrigued by the ln k < k^m). Assuming my first post is "enough" evidence.

Answer 27

agreed. Its actually less work I think because you have to end up integrating by parts with the integral test and taking some limits.

Answer 28

soz agent!

Answer 29

haha jk Irish.

Exactly. I mean it's just like the comparison test, with the p-series http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTests/p-series.html

Answer 30

i was just transfixed by that selfie and lost all sense of reason

Answer 31

you dandy!

Answer 32

I'm just not 100% sure if what I wrote is proof enough for the p-series test, but it seems to be, just by knowing that ln k < k for all k.

And wolframalpha used a comparison test... too bad they no longer allow the three free solutions per day, or else I'd check how they did it.

Answer 33

Someone should sign up for wolram alpha pro so we can check. I would, but... I don't want to.

Answer 34

@legomyego180 when you use the equation tool it automatically puts this `\[` and this `\]` around your math stuff.

If you instead manually type `\(` and `\)` around your math, it will keep it in line with the text, no line break.

Answer 35

Oh cool, thanks zep!

Answer 36

@agent0smith it doesnt offer step by step solutions for series unfortunately

Answer 37

Thank you @zepdrix, \( \text{you are} > angel \)

Answer 38

Nawww Agel Smite is smrt

Answer 39

I always wondered \( \text{how people did that crap} \)

Answer 40

Ya, \(it's\) pretty \(cool\) :D

Answer 41

:D

Then good thing I didn't convince someone to spend their money on WA to see the solution.