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Convergence of the series sin 1/n?

  • 21-11-2011 11:06PM
    #1
    irishdude11
    Registered Users, Registered Users 2 Posts: 377 ✭✭


    Here's how I find the convergence of the series
    Sin 1/n from n = 1 to infinity -

    Take the limit as n -> infinity of Sin 1/n. This gives Sin 0. Sin 0 = 0

    .... But then I remembered that checking if the terms of the sequence approach 0 is only good for testing for divergence, it doesn't imply convergence. So how do you go about testing this series for convergence?


Comments

  • #2
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    Try using the comparison test.

    Note, by the way, that the limit of (sin(x))/x, as x-> 0, is 1. So, you would expect, in the long run, sin(1/n) to behave more and more like 1/n...


  • #3
    Fremen
    Registered Users, Registered Users 2 Posts: 2,481 ✭✭✭Fremen


    Are you referring to the sequence, or the series? A series is a sum of a load of different terms.


  • #4
    irishdude11
    Registered Users, Registered Users 2 Posts: 377 ✭✭irishdude11


    MathsManiac wrote: »
    Try using the comparison test.

    Note, by the way, that the limit of (sin(x))/x, as x-> 0, is 1. So, you would expect, in the long run, sin(1/n) to behave more and more like 1/n...

    Cheers mate. I can see how it works now by letting x = 1/n and taking the limit as x goes to 0.
    Fremen wrote: »
    Are you referring to the sequence, or the series? A series is a sum of a load of different terms.

    I was referring to a series, the terms of the sequence related to the series have to go to 0 otherwise its divergent.


  • #5
    MathsManiac
    Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    irishdude11 wrote: »
    Cheers mate. I can see how it works now by letting x = 1/n and taking the limit as x goes to 0.

    If you're using the limit version of the comparison test, you're more or less sorted, but if you want to use the regular (direct) version, you'll have to think a bit more about the details.


  • #6
    click_here!!!
    Closed Accounts Posts: 119 ✭✭click_here!!!


    MathsManiac wrote: »
    If you're using the limit version of the comparison test, you're more or less sorted, but if you want to use the regular (direct) version, you'll have to think a bit more about the details.

    The limit comparison test works. I found getting the limit tricky though, but got 1 in the end.

    You can also use the integral test, because it's based on a continuous, positive, decreasing function, and that function is integrable by parts.


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