Results 1 to 5 of 5

Math Help - Formal proof of limit/convergence of a sequence

  1. #1
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419

    Formal proof of limit/convergence of a sequence

    Okay, so I have a problem where a_{n}=\frac{1}{n} and I need to show using the formal definition that a_{n} converges to 0:
    \lim_{n\rightarrow \infty}a_{n}=L=0.

    So I let |a_{n}-L| < \varepsilon. Simplifying it, I get
    n > \frac{1}{\varepsilon + L}, essentially getting within distance \varepsilon of L for any integer N where n > N.

    So then I get n > N \ge \frac{1}{\varepsilon + L}.

    Doing some manipulation, I eventually get to \varepsilon > \varepsilon\frac{N}{n}+L\frac{N-n}{n} \ge \frac{1}{n} This is where I get lost on showing that L=0.

    Now my professor didn't really sufficiently explain how to do a formal proof even when we already assume we know the value of L. For instance, we know L=0 for this a_{n}. Knowing this, the inequality would become \varepsilon > \varepsilon\frac{N}{n} \ge \frac{1}{n}. Do we know that this is true because \varepsilon\frac{N}{n} is always less than \varepsilon since \frac{N}{n} < 1, or is there more that needs to be shown? If that is sufficient, than how can that line of thought be applied when we assume we don't know the value of L (like in this problem)?

    Sorry if there is already a thread that asks and explains this. Our class pretty much went through sequences today in about 30 minutes so I'm not too sure on this stuff. Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Pinkk View Post
    Okay, so I have a problem where a_{n}=\frac{1}{n} and I need to show using the formal definition that a_{n} converges to 0:
    \lim_{n\rightarrow \infty}a_{n}=L=0.

    So I let |a_{n}-L| < \varepsilon. Simplifying it, I get
    n > \frac{1}{\varepsilon + L}, essentially getting within distance \varepsilon of L for any integer N where n > N.

    So then I get n > N \ge \frac{1}{\varepsilon + L}.

    Doing some manipulation, I eventually get to \varepsilon > \varepsilon\frac{N}{n}+L\frac{N-n}{n} \ge \frac{1}{n} This is where I get lost on showing that L=0.

    Now my professor didn't really sufficiently explain how to do a formal proof even when we already assume we know the value of L. For instance, we know L=0 for this a_{n}. Knowing this, the inequality would become \varepsilon > \varepsilon\frac{N}{n} \ge \frac{1}{n}. Do we know that this is true because \varepsilon\frac{N}{n} is always less than \varepsilon since \frac{N}{n} < 1, or is there more that needs to be shown? If that is sufficient, than how can that line of thought be applied when we assume we don't know the value of L (like in this problem)?

    Sorry if there is already a thread that asks and explains this. Our class pretty much went through sequences today in about 30 minutes so I'm not too sure on this stuff. Thanks!
    for a given \epsilon > 0, we need to find an integer N > 0 such that for any n \geq N: \ \frac{1}{n}=|a_n| < \epsilon, which is equivalent to n > \frac{1}{\epsilon}. so any integer N > \frac{1}{\epsilon} will work.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    Hmm, so correct me if I'm wrong, so would that imply then that:

    \varepsilon > \frac{1}{n} + L(\frac{1}{\varepsilon n}-1) \ge \frac{1}{n}

    Since \frac{1}{\varepsilon n} - 1 \le 0, L cannot be positive or the inequality does not hold true. But how exactly can we show L is never negative, and therefore L=0.

    Also, doesn't the information you provided already assume L=0? Sorry if these are not the best of questions, but like many others, my teachers and professors have always neglected the formal proofs and definitions (which sucks because I plan on being a pure math major). Thanks again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    normally we don't use the "definition of limit" to "find" the limit. this method is used to "prove" that the limit is what we think it is. for that sequence, our guess was that the limit is 0.

    then i proved it using the definition.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    Ah okay, I thought there was a method to solve for L within the formal definition inequality. Thanks for the help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: August 12th 2011, 07:05 AM
  2. A formal proof for a limit.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2010, 11:18 PM
  3. Alternating sequence, convergence and its limit
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 27th 2010, 03:08 AM
  4. sequence limit and convergence
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 26th 2009, 01:01 PM
  5. Replies: 1
    Last Post: December 2nd 2008, 07:21 AM

Search Tags


/mathhelpforum @mathhelpforum