Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    I have to find if the following series converge:

    Sum from k=1 to k=inf of (1/sqrt(k))*e^(-2*sqrt(k))

    I tried to do it with the ratio test and found that abs(a_(k+1)/a_k) = abs(sqrt(k/(k+1)) * e^ (2*(sqrt(k) * sqrt(k+1))
    which tends to 1 as k tends to infinity, so I can't deduce if the sequence converges or diverges using the ratio test.

    Do you recommend any other way? I don't know with what test to handle it, thanks in advance!
    Online

    18
    ReputationRep:
    Cauchy condensation test should work I think.
    • Thread Starter
    Offline

    0
    ReputationRep:
    We haven't learned this test in our module. Can you give me some more information? I know about comparison test , ratio test (which is not working in this case) and the alternating series test. I don't know what to compare using the comparison test or if the alternating series test is suitable. Thanks in advance for any help!
    Offline

    2
    ReputationRep:
    Integral test, maybe? Havn't got any paper handy, but..

    Function is positive, continuous and decreasing, so it's viable. Maybe.

    _Kar.
    Online

    18
    ReputationRep:
    The comparison test is always "suitable"; however finding a suitable series to compare against (that you can show converges/diverges) may be very hard.

    Ratio/root/alternating series tests definitely don't work.

    It's easy enough to google the Cauchy test.
    Offline

    12
    ReputationRep:
    Unless I'm missing something a comparison test will kill it very quickly : I'm assuming the question is  \frac{1}{\sqrt k} e^{-2\sqrt k} in which case you can focus on the thing that goes smallest (the e) and ignore the 1\sqrt(k)
    Online

    18
    ReputationRep:
    You still need to show \sum e^{-2\sqrt{k}} converges. I felt the quickest way was condensation, but if you have another approach that's fine.
    • Thread Starter
    Offline

    0
    ReputationRep:
    So I will Google about condensation test and try to put it in practice, thank you very very much. If I still have any problems I will write you again! Thank you everyone!
    Offline

    12
    ReputationRep:
    (Original post by DFranklin)
    You still need to show \sum e^{-2\sqrt{k}} converges. I felt the quickest way was condensation, but if you have another approach that's fine.
    Spoiler:
    Show
     exp(x^4)\geq \frac{x^4}{4!} is quick if you permit various properties of exp

    I would have said that doing this is better than using a test the op hasn't learnt but you might disagree and say that the properties I am assuming are too strong
    Online

    18
    ReputationRep:
    (Original post by IrrationalNumber)
    Spoiler:
    Show
     exp(x^4)\geq \frac{x^4}{4!} is quick if you permit various properties of exp

    I would have said that doing this is better than using a test the op hasn't learnt but you might disagree and say that the properties I am assuming are too strong
    A bit 6 of one and half-dozen of the other I guess. In the order I learned analysis you wouldn't have proved this about exp until well after doing all the convergence tests, but that certainly doesn't make what you've done invalid.

    Although I will admit to a sneaking feeling that everyone should learn about the Cauchy condensation test.
    Offline

    8
    ReputationRep:
    (Original post by Darkprince)
    I have to find if the following series converge:

    Sum from k=1 to k=inf of (1/sqrt(k))*e^(-2*sqrt(k))

    I tried to do it with the ratio test and found that abs(a_(k+1)/a_k) = abs(sqrt(k/(k+1)) * e^ (2*(sqrt(k) * sqrt(k+1))
    which tends to 1 as k tends to infinity, so I can't deduce if the sequence converges or diverges using the ratio test.

    Do you recommend any other way? I don't know with what test to handle it, thanks in advance!
    USe the integral test
    This series is monotone decreasing and defined on the [1, \infty[
    open interval so it converges iif
    \displaystyle \int^{\infty}_{1} \frac{1}{\sqrt{x}}e^{-2\sqrt{x}} dx is finite.
    This itegral is a simple case if you consider that with a minus sign it will be the
    \displaystyle \int f'(x)\cdot e^{f(x)} dx integral, of
    which antiderivative is e^{f(x)}
    Online

    18
    ReputationRep:
    In hindsight, the factor of 2 in the exponent looks like it was put there to make the function easy to integrate, so I'd be very surprised if ztibor's method wasn't the one expected.
    Offline

    8
    ReputationRep:
    (Original post by DFranklin)
    In hindsight, the factor of 2 in the exponent looks like it was put there to make the function easy to integrate, so I'd be very surprised if ztibor's method wasn't the one expected.
    It is not the 2 but the \frac{1}{\sqrt{x}} factor make easy to integrate
    because the difference between it and the derivative of \sqrt{x}
    in the exponent is only a constant factor
    Online

    18
    ReputationRep:
    (Original post by ztibor)
    It is not the 2 but the \frac{1}{\sqrt{x}} factor make easy to integrate
    because the difference between it and the derivative of \sqrt{x}
    in the exponent is only a constant factor
    Well yes, it's impossible to integrate without the 1/sqrt(x) term.

    But it's the 2 in the exponent that "doesn't belong there". If you were just looking at convergence, there's "no point" in multiplying by 2. So it looks like it's only there to make it come out nicely when you integrate, and it's a bit of (circumstantial) evidence that integration was the intended method.
    • Thread Starter
    Offline

    0
    ReputationRep:
    I did it using the comparison method, thanks for your replies
 
 
 
Poll
Do you think parents should charge rent?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.