Results 1 to 6 of 6
Like Tree2Thanks
  • 1 Post By romsek
  • 1 Post By Deveno

Math Help - Real Analysis Question

  1. #1
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    424
    Thanks
    11

    Real Analysis Question

    Sorry if this is in the wrong section, please move it if it is.

    Suppose that ${y_n}$ is a sequence of real numbers. Then $y_n \rightarrow \infty $ as $n \rightarrow \infty $ means that for every $M \in \mathbb{R}$, there exists $N \in \mathbb{N} $ such that
    $n \geq N \Rightarrow y_n > M$

    Suppose that $x_n > 0$ for all $n$ and $x_n \rightarrow 0$ as $n \rightarrow \infty$. Prove that $\dfrac{1}{x_n} \rightarrow \infty$ as $n \rightarrow \infty$
    So I'm not very good with this real analysis stuff, this whole new definition of a convergent sequence has thrown me off.

    Since $x_n \rightarrow 0$ as $ n \rightarrow \infty$ there exists $N \in \mathbb{N}$ for every $M \in \mathbb{R}$ such that $ n \geq N \Rightarrow x_n < M$

    I've pretty much just copied the first line of the question, but I don't know where to start... can anyone point me in the right direction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,225
    Thanks
    851

    Re: Real Analysis Question

    Let $y_n=\dfrac{1}{x_n}$

    We want to show that $\displaystyle{\lim_{n\to \infty}}x_n=0 \Rightarrow \displaystyle{\lim_{n\to \infty}}y_n=\infty$

    Let $M>0$

    Since $\displaystyle{\lim_{n\to \infty}}x_n=0$ we know $\exists N \in \mathbb{N} \ni n>N \Rightarrow x_n<\dfrac{1}{M}$

    and thus $n>N \Rightarrow y_n=\dfrac{1}{x_n}>M$ and thus $\forall M>0 ~\exists N \in \mathbb{N} \ni y_n>M$

    and thus $\displaystyle{\lim_{n\to \infty}}y_n=\infty$
    Thanks from Educated
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    424
    Thanks
    11

    Re: Real Analysis Question

    Wow thank you so much. It seems simple now that you've done it like that. I'm still trying to fully grasp the concept of this epsilon delta definition of limits and sequences.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,316
    Thanks
    697

    Re: Real Analysis Question

    Here is what "epsilon/delta" means in "ordinary language":

    The formal definition goes:

    $\displaystyle \lim_{n \to \infty} x_n = L$ if:

    for ANY $\epsilon > 0$, there is a positive integer $N$ such that:

    for ALL $n > N: |x_n - L| < \epsilon$.

    Now, the thing to realize is that we're actually hoping to prove this for VERY SMALL epsilons, because if epsilon is HUGE, it might be that ANY $N$ works.

    Of course, it might take a lot of terms for $x_n$ to "settle down", so we should expect that $N$ will be rather large. If a certain $N$ doesn't work, we look for a "bigger one".

    When $L$ is not finite, we have to take a new tack, because infinity is NOT a real number, so we cannot set any other real number "equal" to it. But what do we mean when we say something is (positive, yet) infinite?

    We mean that it is larger than ANY real number. Formally, this becomes:

    $\displaystyle \lim_{n \to \infty} x_n = \infty$ if:

    for any $M > 0$, we can find SOME positive integer $N$ with:

    $x_n > M$ for ALL $n > N$. Again, we might have to choose $N$ quite large to ensure every term $x_n$ that comes after $x_N$ is larger than $M$.

    A word about why we want $x_n > 0$ for all $n$. If not, then $\dfrac{1}{x_n}$ might swing violently up and down, and we can't say that "converges" in any meaningful sense of the word. In fact, such a sequence might eventually take on all values of any countable subset of the reals.

    As romsek's answer shows, if we are GIVEN the $N$ to fulfill the "0-limit case" (here, he is using the positive number $\dfrac{1}{M}$ in place of "epsilon" which is fine, since $M$ may be large, its reciprocal may be very small) we can use that SAME $N$ for the "reciprocal sequence" (if $\epsilon$ is arbitrarily small, then $1/\epsilon$ is arbitrarily large).
    Thanks from Educated
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Educated's Avatar
    Joined
    Aug 2010
    From
    New Zealand
    Posts
    424
    Thanks
    11

    Re: Real Analysis Question

    So for some reason, our class is really pedantic about this sort of stuff. I asked my tutor if romsek's answer was correct, but he said no (I guess that's not how he wants us to write it out)... but it makes full sense to me.

    I attempted it myself, with his aid and came up with this solution:

    Let $M \in \mathbb{R}$. We want to find $N \in \mathbb{N}$ such that $ n \geq N \Rightarrow \frac{1}{x_n} >M$

    Since $x_n \rightarrow 0$ as $n \rightarrow \infty$, there exists $N \in \mathbb{N}$ such that
    $ n \geq N \rightarrow |x_n - 0| < \frac{1}{M}$

    Because $x_n > 0$ we have:
    $n \geq N \Rightarrow x_n < M \Rightarrow \frac{1}{x_n} > M$
    and so this N works.

    But he said it can be better. What more must I add? Everything seems to be in order.

    It seems like the idea is the same, but they just want me to write it in a specific way...


    And Deveno, it takes me a while to fully understand the concept. I kind of get it, but not fully. I remember when I kept asking, "How does this prove that this sequence converges to x as n goes to infinity?" but I sort of get why.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,316
    Thanks
    697

    Re: Real Analysis Question

    Well, what do we MEAN when we say a sequence "converges"? We mean it gets close to a certain value, and moreover, it keeps getting closer as we take "further terms".

    How do we express the fact that two real numbers are close? We want the DISTANCE between them to be small. What is the distance between a and b? It is |a - b| = |b - a|.

    So when we say:

    $|x_n - L| < \epsilon$, what we mean, informally, is that $x_n$ is "within epsilon" of $L$, or more precisely, that $x_n$ lies in the (small) interval $(L -\epsilon,L+\epsilon)$.

    All of this still holds when $L = 0$, although some of the expressions are now a bit simpler.

    I don't know what your tutor's problem is.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question in real analysis
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 18th 2013, 09:20 AM
  2. Real Analysis Question
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 28th 2011, 03:45 PM
  3. One Question In Real Analysis ..
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 24th 2010, 01:51 PM
  4. real analysis question
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 20th 2008, 10:47 PM
  5. real analysis question
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 20th 2008, 07:56 PM

Search Tags


/mathhelpforum @mathhelpforum