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Homework Help: Convergent sequence

  1. Oct 31, 2015 #1
    1. The problem statement, all variables and given/known data
    Show, from the definition of what it means for a function to converge to a limit, that the sequence ##\left\{x^t\right\}_{t=1}^{\infty}## with ##x^t = \frac{2t+5}{t^2+7}## converges to ##0## as ##t## goes to infinity.

    2. Relevant equations
    A sequence converges to ##x^0 \in X## if for any ##\epsilon > 0##, there is ##N \in \mathbb{N}## such that if ##t > N##, then ##d(x^t,x^0) < \epsilon##.

    3. The attempt at a solution
    To show that ##x^t = \frac{2t+5}{t^2+7}## converges to ##0## we must, for any ##\epsilon > 0##, find a value ##N## such that if ##t > N##, then
    $$\left|\frac{2t+5}{t^2+7} - 0\right| = \left|\frac{2t+5}{t^2+7}\right| < \epsilon$$.
    Now sure how to simplify ##x^t## to show that it is less than or equal to some much simpler expression in ##t## that can clearly be made less than any given ##\epsilon## by choosing ##n## large enough. Please help.
     
  2. jcsd
  3. Oct 31, 2015 #2

    PeroK

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    First, you could note that ##t^2 +7 > t^2##
     
  4. Nov 1, 2015 #3

    HallsofIvy

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    Since t is positive so is the fraction so you can drop the absolute value sign: [itex]\frac{2t+ 5}{t^2+ 7}< \epsilon[/itex] so [itex]2t+ 5< \epsilon(t^2+ 7)[/itex]. For what t is [itex]\epsilon t^2- 2t+ 7\epsilon- 5> 0[/itex]?
     
  5. Nov 1, 2015 #4

    Ray Vickson

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    If you do not want the smallest possible ##N## giving ##|x_t| < \epsilon## for ## t > N##, but are satisfied with some ##N = N(\epsilon)## that 'works', the problem is quite straightforward. Do you agree that for ##t \geq 1## the numerator is ##\leq 7t##? Do you agree that the denominator is ##> t^2##?
     
  6. Nov 1, 2015 #5
    Yup that's obvious so is it as easy as saying ##\frac{2t}{t^2} = \frac{2}{t} < \epsilon##.
     
  7. Nov 1, 2015 #6

    Ray Vickson

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    Well, no: ##5 + 2 \neq 2##.
     
  8. Nov 1, 2015 #7
    Then I'm a little stuck. That ##t^2## has lost me. It's easy if it's just ##t##.
     
  9. Nov 1, 2015 #8

    Ray Vickson

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    ##\frac{7t}{t^2} = \frac{7}{t}##. Note that the original numerator is ##7 t##, not ##2 t## as you wrote (but otherwise you did exactly the same simplification, using 2 instead of 2+5 = 7).
     
  10. Nov 1, 2015 #9
    Ok why is the numerator ##\leq 7t## for ##t > 0##??
     
  11. Nov 1, 2015 #10

    Ray Vickson

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    It isn't, and I never claimed it was. Go back and read what I wrote, word-for-word.
     
  12. Nov 1, 2015 #11
    So is it as easy as saying ##\frac{7t}{t^2} = \frac{7}{t} < \epsilon##??
     
    Last edited: Nov 1, 2015
  13. Nov 1, 2015 #12
    EDIT:
    Only if you understand how you got there...



    Solve the two inequalities in terms of ##t##:

    ##2t + 5 \leq 7t##

    and

    ##\frac{1}{t^2 + 7} \leq \frac{1}{t^2}##

    What relationship do you get for t in each one? What does ##\left\{x^t\right\}_{t=1}^{\infty}## dictate for t?

    If ##A \leq B## and ##B < C##, then ##A < C##, yes?
     
    Last edited: Nov 1, 2015
  14. Nov 1, 2015 #13
    For the first one ##t \geq 1## and second one ##t > 0##. Where am I going with this??
     
  15. Nov 2, 2015 #14
    So ##\frac{2t+5}{t^2+7} \leq \frac{7t}{t^2} = \frac{7}{t}## for ##t > 0##??
     
  16. Nov 2, 2015 #15
    Perfect! That's very good. (except it's for ##t \geq 1##, look at post #13 again... And our sequence specifies valid values for t (also t > n, which is a natural number...). Is our inequality ever violated?)

    And if we can show ##\frac{7}{t} < \epsilon##, then we know that ##\frac{2t+5}{t^2+7} < \epsilon##, right?

    If ##n \in \mathbb{N}##, is ##(7 \times n) \in \mathbb{N}##, as well?
    If ##t > n## does it hold that ##7 \times t > 7 \times n##?

    (Note: I will use lowecase n because I don't want to make it confusing between the variable and the set)
     
    Last edited: Nov 2, 2015
  17. Nov 2, 2015 #16
    Sorry, you were wrong in #13 and I didn't catch it. It's a minor error, but

    ##\frac{1}{t^2 + 7} \leq \frac{1}{t^2}##

    ##t^2 \leq t^2+7##

    ## 0 \leq 7##

    thus, the second inequality is always true.
     
  18. Nov 2, 2015 #17

    PeroK

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    One thing you're missing is that you don't need inequalities to hold for all ##t > 0##. If you are taking the limit ##t \rightarrow \infty## then you can look for inequalities that hold for large enough t. For example,

    For ##t > 5## we have ##2t + 5 < 3t##
     
    Last edited: Nov 2, 2015
  19. Nov 2, 2015 #18
    So ##t\geq 1##.
     
  20. Nov 2, 2015 #19
    Not sure what u mean here.
     
  21. Nov 2, 2015 #20
    I chose ##N > \frac{7}{\epsilon}## and we shall have if that ##t > N##, then $$\left|\frac{2t+5}{t^2+7}\right| < \epsilon.$$
     
  22. Nov 2, 2015 #21

    PeroK

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    What you are trying to do with this problem is show what happens to your function as ##t \rightarrow \infty##. This means that when you are working with your function, you can assume that ##t## is greater than a given value. Any concerns about what happens if ##t = 0## or ##t = 1## are irrelevant to the limit. Let me given you an example.

    Suppose I wanted to show that ##t^3 + 6t^2 - 15t + 173 \rightarrow \infty## as ##t \rightarrow \infty##

    Well, I could try to solve cubic equations. But, instead, I would do something like this:

    For ##t > 0, \ t^3 + 6t^2 - 15t + 173 > t^3 - 15t = t(t^2 - 15)##

    For ##t > 4, \ t(t^2 - 15) > t##

    Hence, for ##t > 4, \ t^3 + 6t^2 - 15t + 173 > t##

    And, then it's very easy.
     
  23. Nov 2, 2015 #22
    Ok thanks!!
     
  24. Nov 2, 2015 #23
    I think you might be caught up on the wrong things. We are working backwards along the following proof. Normally I wouldn't post an entire proof, but you've already solved the problem. I want to make sure you have a complete understanding of how that solution worked.

    1. There exists a natural number ##a##, such that ##x > a## satisfies the inequality: ##\frac{1}{x} < \epsilon## This is known/obviously true because ##\lim_{x \to ∞} \frac{1}{x} = 0 < \epsilon##
    2. ##\frac{7}{7x} < \epsilon## 1
    3. ##\{7, a\} \in ℕ## 1 and definition of ℕ
    4. ##7a \in ℕ## 3 and closure of ℕ for multiplication
    5. let ##n = 7a##
      let ##t = 7x##
    6. ##7x > 7a## 1
      ##t > n## 1,5 and 7x > 7a
    7. ##n \in ℕ## 4,5
    8. ##\frac{7}{t} < \epsilon## 2, 5
      ##\frac{7t}{t^2} < \epsilon##
      To recap: We have shown that there exists a number "n" in ℕ, and that if t > n, then ##\frac{7}{t} < \epsilon##. The rest if the proof is algebra to show that ##\frac{5+2t}{t^2+7} < \frac{7}{t} < \epsilon##
    9. ##7 > 0## axiom
      ##t^2 + 7 > t^2##
      ##\frac{1}{t^2} > \frac{1}{t^2+7}##
      ##\frac{7t}{t^2} > \frac{7t}{t^2+7}##
    10. ##t > 1## This is a new constraint on our solution that t is greater than 1, but it doesn't matter because we can make n (and therefore t) arbitrarily large
      ##5t > 5##
      ##7t > 5+2t##
      ##\frac{7t}{t^2+7} > \frac{5+2t}{t^2+7}##
    11. ##\frac{5+2t}{t^2+7} < \frac{7t}{t^2+7} < \frac{7t}{t^2} < \epsilon## 10, 9, 8

    ∴ ##\frac{5+2t}{t^2+7} < \epsilon## 11, and ##t > n## 6, and ##n \in ℕ## 7
     
    Last edited: Nov 2, 2015
  25. Nov 2, 2015 #24

    Ray Vickson

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    You were the one that said ##\{x_t\}_{t=1}^{\infty}## in post # 1, so that means you want ##t = 1, 2, 3, \ldots##. In other words, you only care about ##t \geq 1##. But, in fact, to analyze what happens as ##t \to \infty## you could assume ##t > 1,000,000## if you wanted to---because eventually you will be taking ##t## larger than any number you can name or write down. That's what ##t \to \infty## means.

    So, ##t \geq 1##, agreed? Now, for ##t \geq 1## we have ##5 \leq 5 t##, so ##2t + 5 \leq 2t + 5t = 7t##. That's all there is to it: just basic properties of numbers.

    You also have ##t^2 + 7 > t^2##, so for ##t \geq 1## you have
    [tex] 0 < \frac{2t+5}{t^2+7} < \frac{7t}{t^2} = \frac{7}{t} [/tex]
     
  26. Nov 2, 2015 #25
    If you don't like the given sequence, you can work the problem on a simple equivalent in a neighborhood of ##+\infty##.
     
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