Question

I have a series convergence question: Find the values of the parameter p for which the following series converges:

Answer 1

\[\sum_{k=2}^{\infty} \frac{ 5 }{ (4\ln(\ln k))^{p}) }\]

Answer 2

hmmm...what tests have you studied?

Answer 3

divergence, ratio, root, limit comparison...

Answer 4

for some reason I'm thinking we should try a comparison test

Answer 5

hmm...i'll try that

Answer 6

\[\log(\log(k))<k \text{ for } k>1 \]

Answer 7

I have a hunch that the ratio test might be helpful... not sure yet, working on the result.

Answer 8

just to clarify the entire bottom is being raised to the pth power correct?

Answer 9

yes

Answer 10

k

Answer 11

im not sure...my results for comparison test says the series diverges...=/ stuck

Answer 12

for which p?

Answer 13

I don't think it should diverge for all p

Answer 14

could be wrong though

Answer 15

\[\frac{1}{k}< \frac{1}{\ln(\ln(k))} \\ \text{ if } p>0 \text{ then we have } \\ \frac{1}{k^p}< \frac{1}{(\ln(\ln(k)))^p} \\ \]
so in this comparison the only thing we can see if for p>0 we have divergence

Answer 16

\[\int\limits_{2}^{\infty} \frac{1}{k^p} dk =\int\limits_2^\infty k^{-p} dk \\ \text{ assume } p \neq 1 \text{ then we have } \frac{k^{-p+1}}{-p+1}|_2^{\infty} \\ =\lim_{z \rightarrow \infty}(\frac{z^{1-p}}{1-p})-\frac{2^{1-p}}{1-p} \\ \\ \text{ hmm if } 1-p>0 \text{ then the \integral diverges } \\ \text{ and if } 1-p<0 \text{ then \integral converges } \]

but we can only use the diverging thing since we chose a sequence less than the given sequence for comparison
so the only thing we have shown is that for 1-p>0 we have the series diverges

Answer 17

oops and we had p>0 above
so actually p is between 0 and 1 there

Answer 18

but still need to look at other p cases

Answer 19

for example what happens at p=0 and p=1
and everything outside the (0,1) interval

Answer 20

so the only thing I have shown is this:
\[\text{ Since where } 0<p<1 \text{ the } \int\limits _2^\infty \frac{1}{k^p} dk \text{ diverges then } \\ \sum_{n=2}^{\infty} \frac{1}{k^p} \text{ diverges and since } \\ \text{ for } 0<p<1 \text{ we have } \frac{1}{(\ln(\ln(k))^p}> \frac{1}{k^p} \text{ then } \sum_{n=2}^{\infty} \frac{1}{(\ln(\ln(k))^p} \\ \text{ diverges again for } 0<p<1 \\ \text{ we still need to look at other } p \\ \text{ also any constant value times that want change the divergence except maybe 0 :p}\]

Answer 21

won't (not want)

Answer 22

if p=1 above..
\[\int\limits_{2}^{\infty}\frac{1}{k} dk=\ln|k|_2^\infty \]
this still diverges

Answer 23

so you could say 0<p<=1 we have divergence

Answer 24

so I think we need another sequence for comparision for p>1
we also need to consider p=0
and p<0
brb...

Answer 25

Clearly, p < 0 diverges as this will bring the denominator on top... and p = 0 gives an infinite series of 5's also divergent.

Answer 26

true above the p=0 thing ...
and yes for p<0 we should have divergence

Answer 27

so @cstarxq looks like you were correct for all p this thing diverges

Answer 28

wait we didn't consider p>1 yet

Answer 29

thanks, i just finished this hw question online, the series diverges for all values of p

Answer 30

oh ok cool

Answer 31

was it multiple choice?

Answer 32

with fill-ins

Answer 33

oh ok
cool stuff

Answer 34

still trying to think what we can do for p>1

Answer 35

I know you already done it
but I might play with a bit more

Answer 36

:) that's cool

Answer 37

@jtvatsim do you have anything for p>1 ?

Answer 38

Still working on it... Mathematica says to use a comparison test, but doesn't specify what to compare it to.

Answer 39

I was wondering if maybe something like
1/ln(ln k)^p > 1/ln(k)^p
might be helpful if we somehow express p in terms of an "e" to get rid of the logs...

Answer 40

i actually tried that, getting rid of the logs...but got stuck on that

Answer 41

for limit comparison test, compare it to 1/k?

Answer 42

tried that... lim comp fails... goes to infinity.

Answer 43

Ah... might have something.

Answer 44

This should show that the series diverges for p >= e.

Given that ln(ln k) < ln k < k for all k > 1. It follows that
1/ln(ln k) > 1/ln k > 1/k for all k > 1. Furthermore, for p > 0 we have,
(1/ln(ln k))^p > (1/ln k)^p > (1/k)^p.

Focus on the first inequality, namely,
(1/ln(ln k))^p > (1/ln k)^p
and suppose that p >= e. Then, p = e + r where r >= 0.

Then,
(1/ln(ln k))^p > (1/ln k)^(e + r) = 1/[(ln k)^e (ln k)^r] = 1/(ln k)^r * 1/k.
But this is just a positive constant times a divergent series. Hence, by comparison
we have shown that the series diverges for all p >= e.

A similar argument shows that the series diverges for 0 < p <= e as well.

Answer 45

are you saying:
\[\frac{1}{(\ln(k))^e}=\frac{1}{k}?\]

Answer 46

dang it... lol

Answer 47

Assume (ln k)^e = 1. Then the result follows. LOL. That's called seeing what you want to. :)

Answer 48

I totally get that. I do that sometimes.

Answer 49

Then everyone else is like math is really magic.

Answer 50

Yep, 0 = 1 implies anything.

Answer 51

I'm kind of stumped

Answer 52

Yeah, this is a tough question. There must be some trick involved here.

Answer 53

It said that the limit comparison for harmonic series is unbounded for all values of p, which means that the series diverges for all values of p... ?

Answer 54

Really? Cuz, that didn't work for me. The way my math is going right now though, I guess that may be true...

Answer 55

Well, I am done for tonight. Have a good one everyone! :)

Answer 56

goodnight, thanks for helping!~

Answer 57

goodnight @jtvatsim

Answer 58

@cstarxq sadly I guess I'm not picking the right series for comparison
I give up for tonight :(

Answer 59

oh i think i see I guess i forgot all the parts of the comparison limit test

Answer 60

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&cad=rja&uact=8&ved=0CDIQFjAD&url=http%3A%2F%2Ffaculty.bucks.edu%2Ferickson%2Fmath141%2F141chap9.pdf&ei=AXGTVZqQGsiCsAWszoDgCw&usg=AFQjCNH8TAWi7iA3eNJKKQAQN1hHtM_F7A&sig2=ZbCUCOFfBhLahryJWPpNVA page 19
...
so I was looking at:
\[\lim_{k \rightarrow \infty} \frac{1}{\ln(\ln(k))} \div \frac{1}{k} =\infty\]
which we can raise both sides by p
and we know
as you said the harmonic series diverges for all p
so the other series also diverges for all p according to page 19
thereom 9.36 part 3

Answer 61

http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx I was using pauls notes and did not see this for the limit comparison test.

Answer 62

That was my guide until I did some more searching.

Answer 63

and found that pdf

Answer 64

@myininaya oh wow, that pdf will be really helpful for studying for my upcoming exam. Thanks for sharing!