Results 1 to 8 of 8

Math Help - Convergence of sequence using definition

  1. #1
    Member
    Joined
    Dec 2012
    From
    -
    Posts
    77

    Convergence of sequence using definition

    "using the definition (epsilon-N definition), show that the sequences (-1)^n(1/3n) + (1/2n) and cos(n^2)/(3n+1) both converge to the limit zero". ok, so i know the limit and i know the process. i can handle these questions ok when it is just a polynomial sequence but, for example, what do i do with the (-1)^n? the first step i need to do is deal with this and i don't know how to proceed. similar for cosines.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,318
    Thanks
    1234

    Re: Convergence of sequence using definition

    Can you show us what you tried please?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2012
    From
    -
    Posts
    77

    Re: Convergence of sequence using definition

    hi,
    as i said, i encountered a problem with the first step. if someone can tell me how to deal with the (-1)^n, i can do the rest. but i cant try anything beyond that because the obstacle i need help with is right at the beginning..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1

    Re: Convergence of sequence using definition

    Quote Originally Posted by learning View Post
    hi,
    as i said, i encountered a problem with the first step. if someone can tell me how to deal with the (-1)^n, i can do the rest. but i cant try anything beyond that because the obstacle i need help with is right at the beginning..
    For the first one separate it.
    \left| {{{( - 1)}^n}{{\left( {\frac{1}{3}} \right)}^n} + \frac{1}{{2n}}} \right| \le \left| {{{( - 1)}^n}{{\left( {\frac{1}{3}} \right)}^n}} \right| + \left| {\frac{1}{{2n}}} \right|

    Make each of those terms less than \frac{\varepsilon }{2}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2012
    From
    -
    Posts
    77

    Re: Convergence of sequence using definition

    hmmm its 1/3n, not (1/3)^n. so does this make a difference? my thought was that i should just say that the whole thing is less than (something?) regardless of sign so i.e. take the value for even n and say its less than that?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Convergence of sequence using definition

    To prove \lim_{n \to \infty} (-1)^n \frac{1}{3n}+\frac{1}{2n}=0, i.e \forall \epsilon>0, \exists P>0: |n|>P \Rightarrow \left|(-1)^n \frac{1}{3n}+\frac{1}{2n}\right|<\epsilon
    Proof:
    Let \epsilon>0 then we have
    \left|(-1)^n \frac{1}{3n}+\frac{1}{2n}\right| = \left|\frac{1}{3n}\right|+\left|\frac{1}{2n}\right  |=\left|\frac{1}{n}\right|\frac{5}{6}.
    Since |n|>P \Rightarrow \left|\frac{1}{n}\right|<\frac{1}{P}.

    Can you finish the proof?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2012
    From
    -
    Posts
    77

    Re: Convergence of sequence using definition

    what i did is:

    |(-1)^n\frac{1}{3n}+\frac{1}{2n}|=|\frac{(-1)^n2n + 3n}{6n^2}| \leq |\frac{5n}{6n^2}| \leq |\frac{5}{6n}| etc.... is this the same/an ok way of expressing it? basically saying that 'it has to be less than what it is if n is even', as that gives the highest absolute value. i mean its technically true, do i have to be more rigid in the conditions here? it leads to same the resut. and yes i am confortable with the rest of the proof it was just how to deal with the (-1)^n i have never come across. basically the differnce is i put the two parts over a common denomnator instead of splitting. in fact once the (-1)^n is gone i can just get rid of the modulus altogether since its all always positive then.
    Last edited by learning; March 24th 2013 at 01:53 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Convergence of sequence using definition

    Can you give the proof for the other limit now?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 18
    Last Post: April 12th 2012, 02:34 PM
  2. definition of convergence of a sequence
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 23rd 2011, 11:04 AM
  3. Convergence of sequence using the definition
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: August 2nd 2010, 07:47 AM
  4. Definition of a sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 27th 2010, 05:10 PM
  5. Replies: 6
    Last Post: October 24th 2008, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum