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

- anonymous

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- schrodinger

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- anonymous

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

- myininaya

hmmm...what tests have you studied?

- anonymous

divergence, ratio, root, limit comparison...

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## More answers

- myininaya

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

- anonymous

hmm...i'll try that

- myininaya

\[\log(\log(k))1 \]

- jtvatsim

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

- jtvatsim

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

- anonymous

yes

- jtvatsim

k

- anonymous

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

- myininaya

for which p?

- myininaya

I don't think it should diverge for all p

- myininaya

could be wrong though

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

but still need to look at other p cases

- myininaya

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

- myininaya

so the only thing I have shown is this:
\[\text{ Since where } 0

\frac{1}{k^p} \text{ then } \sum_{n=2}^{\infty} \frac{1}{(\ln(\ln(k))^p} \\ \text{ diverges again for } 0

- myininaya

won't (not want)

- myininaya

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

- myininaya

so you could say 0

- myininaya

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

- jtvatsim

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

- myininaya

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

- myininaya

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

- myininaya

wait we didn't consider p>1 yet

- anonymous

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

- myininaya

oh ok cool

- myininaya

was it multiple choice?

- anonymous

with fill-ins

- myininaya

oh ok
cool stuff

- myininaya

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

- myininaya

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

- anonymous

:) that's cool

- myininaya

@jtvatsim do you have anything for p>1 ?

- jtvatsim

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

- jtvatsim

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...

- anonymous

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

- anonymous

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

- jtvatsim

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

- jtvatsim

Ah... might have something.

- jtvatsim

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.

- myininaya

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

- jtvatsim

dang it... lol

- jtvatsim

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

- myininaya

I totally get that. I do that sometimes.

- myininaya

Then everyone else is like math is really magic.

- jtvatsim

Yep, 0 = 1 implies anything.

- myininaya

I'm kind of stumped

- jtvatsim

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

- anonymous

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... ?

- jtvatsim

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

- jtvatsim

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

- anonymous

goodnight, thanks for helping!~

- myininaya

goodnight @jtvatsim

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

That was my guide until I did some more searching.

- myininaya

and found that pdf

- anonymous

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

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