- anonymous

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

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.