Bounded Monotonic Sequence Theorem

  • #1
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3

Homework Statement


[/B]
Use the Bounded Monotonic Sequence Theorem to prove that the sequence:

[tex] \{a_{i} \} = \Big\{ i - \sqrt{i^{2}+1} \Big\} [/tex]

Is convergent.


Homework Equations




The Attempt at a Solution


[/B]
I've shown that it has an upper bound and is monotonic increasing, however it is to my understanding that for me to use this theorem the sequence must be bounded (and of course have monotonicity) - i.e. have an upper and lower bound. I can't seem to show that this has a lower bound; even graphically, it just blows off to negative infinity. How should I proceed?
 

Answers and Replies

  • #2
34,521
6,212

Homework Statement


[/B]
Use the Bounded Monotonic Sequence Theorem to prove that the sequence:

[tex] \{a_{i} \} = \Big\{ i - \sqrt{i^{2}+1} \Big\} [/tex]

Is convergent.


Homework Equations




The Attempt at a Solution


[/B]
I've shown that it has an upper bound and is monotonic increasing, however it is to my understanding that for me to use this theorem the sequence must be bounded (and of course have monotonicity) - i.e. have an upper and lower bound. I can't seem to show that this has a lower bound; even graphically, it just blows off to negative infinity.
Why do you think this? You didn't say anything about the possible values of i, but I assume they are positive integers {1, 2, 3, ...}.

Consider ##f(x) = x - \sqrt{x^2 + 1}##. All of the values of this function are negative, since ##x < \sqrt{x^2 + 1}## for all x > 0. As x gets larger, the difference is smaller, and approaches zero. The largest difference comes when x is smallest (i.e., closest to zero).
Morgan Chafe said:
How should I proceed?
 
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  • #3
Ray Vickson
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Homework Statement


[/B]
Use the Bounded Monotonic Sequence Theorem to prove that the sequence:

[tex] \{a_{i} \} = \Big\{ i - \sqrt{i^{2}+1} \Big\} [/tex]

Is convergent.


Homework Equations




The Attempt at a Solution


[/B]
I've shown that it has an upper bound and is monotonic increasing, however it is to my understanding that for me to use this theorem the sequence must be bounded (and of course have monotonicity) - i.e. have an upper and lower bound. I can't seem to show that this has a lower bound; even graphically, it just blows off to negative infinity. How should I proceed?
If ##f(i) = i -\sqrt{i^2+1}## is monotonically increasing, then ##f(1) < f(2) < f(3) < \cdot##, so if ##f(1)## is a finite number, it is a lower bound!
 
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Likes Euler2718
  • #4
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3
Why do you think this? You didn't say anything about the possible values of i, but I assume they are positive integers {1, 2, 3, ...}.

Consider ##f(x) = x - \sqrt{x^2 + 1}##. All of the values of this function are negative, since ##x < \sqrt{x^2 + 1}## for all x > 0. As x gets larger, the difference is smaller, and approaches zero. The largest difference comes when x is smallest (i.e., closest to zero).

If ##f(i) = i -\sqrt{i^2+1}## is monotonically increasing, then ##f(1) < f(2) < f(3) < \cdot##, so if ##f(1)## is a finite number, it is a lower bound!

Thank you, I think I got it now.
 

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