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Prove that a monotone increasing and bounded sequence converges

  1. Apr 29, 2014 #1
    If ##\left\{ a_{n} \right\}## is monotone increasing and there exists ##M \in \Re## such that for every ##n \in N## ##a_{n} ≤ M## prove that ##\left\{ a_{n} \right\}## converges. (Hint: Use the Cauchy sequence property.

    Recall:
    1) ##\left\{ a_{n} \right\}## is Cauchy if and only if ##\left\{ a_{n} \right\}## converges
    2) A sequence is Cauchy if for any ##\epsilon > 0## there exists an ##N## such that for every ##i,j \in \aleph ≥ N## it follows that ##\left| a_{i} - a_{j} \right| < \epsilon##.
    3) A sequence is called monotone increasing if for every ##i < j ## ##a_{i} ≤ a_{j}##.

    Before I attempt to offer a formal proof let me mention that I am getting stuck. Because of the hint that is provided I believe it should be easier to prove that the sequence ##\left\{ a_{n} \right\}## is Cauchy (which is equivalent to proving that it converges). Keeping this in mind let me outline my approach:

    1) Assume that ##\left\{ a_{n} \right\}## is monotone increasing and bounded above by M. (The condition that ##a_{n} ≤ M## for all ##n \in N## is equivalent to being bounded above, correct? I mean...that's the definition, right?)
    2) Since ##\left\{ a_{n} \right\}## is bounded above there exists a least upper bound which I will call ##L##. Intuitively I suspect that ##\left\{ a_{n} \right\}## will converge to ##L##. So I have the condition ##L ≤ M##.
    3) Since ##\left\{ a_{n} \right\}## is monotone increasing I know for every ##i < j ## ##a_{i} ≤ a_{j} ≤ L ≤ M##.

    4) Let ##\epsilon > 0## be given. We must find an ##N## such that for any ##i, j ≥ N## it follows that ##\left| a_{i} - a_{j} \right| < \epsilon## (this is the definition of a Cauchy sequence which will be sufficient to prove that the sequence converges).
    5) There is an ##N## for which ##L - \epsilon < a_{N} ≤ L##. Then we take our ##i, j ≥ N## to arrive at the following two inequalities:

    $$L - \epsilon < a_{N} ≤ a_{i} ≤ L$$
    $$L - \epsilon < a_{N} ≤ a_{j} ≤ L$$

    Or, perhaps more succinctly,

    $$L - \epsilon < a_{i} ≤ L$$
    $$L - \epsilon < a_{j} ≤ L$$


    I'm not sure where to go from here. I feel as if I'm on the right track but then I just get stuck.

    Another qualm I have is that we didn't discuss the notion of a least upper bound in class and so I suspect that this problem can be approached with using the LUB.


    I would appreciate any help pushing me forward with what I have so far and also any help with another way to prove this with the LUB (if possible).
     
  2. jcsd
  3. Apr 29, 2014 #2

    mfb

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    ##L - \epsilon < a_{N}## can be written as ##L < a_{N} + \epsilon##
    Then combine it with ##a_{i} ≤ L## to find a new inequality between ##a_N## and ##a_i##.
     
  4. Apr 29, 2014 #3

    Fredrik

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    These equalities should tell you how to find something greater than ##|a_i-a_j|##. You can consider the cases ##i>j## and ##j>i## separately to eliminate any issues caused by the absolute value.

    I was wondering about this as I read the problem statement. If we use the fact that there's a least upper bound L, then I find it easier to prove that the sequence converges to L than to prove that it's a Cauchy sequence. The strategy is to show that if it doesn't converge to L, there's an upper bound less than L. But after a few minutes of thought, I still don't see a way to show that it's a Cauchy sequence without using that there's a least upper bound.
     
  5. Apr 29, 2014 #4
    Alright, so I know that ##L - \epsilon < a_{N} ≤ L## or ##L - \epsilon < a_{N}## so I can write ##L < a_{N} + \epsilon##. We also know that ##a_{i}## and ##a_{j}## are both less than or equal to ##L## and ##i<j## so I can write ##a_{i} ≤ a_{j} ≤ L < a_{N} + \epsilon## so I have the inequalities

    $$a_{i} < a_{N} + \epsilon$$
    $$a_{j} < a_{N} + \epsilon$$

    Where should I go from here? I can subtract ##a_{N}## from both sides of each of these inequalities but then what?


    I'm not sure how to find something greater than ##|a_i-a_j|##. If I took my ##a_{i}## and ##a_{j}## and subtracted them in the inequalities I would be left with ##0 < a_{i} - a_{j} ≤ 0## which really doesn't seem useful to me.

    That's exactly why I'm a little bit confused about this. Maybe I'm missing something that would make this work? Any other thoughts? Or is the hint not really so useful here?
     
  6. Apr 29, 2014 #5

    jbunniii

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    If the sequence is not Cauchy, then there is some ##\epsilon > 0## such that for every ##N##, we can find some ##n > m \geq N## with ##a_n - a_m > \epsilon##. We can use this to construct a subsequence which diverges to ##+\infty##, and then monotonicity implies that the sequence itself must also diverge to ##+\infty##.

    But I agree that it's more straightforward to prove directly that ##a_n## converges to the least upper bound ##L##.
     
  7. Apr 29, 2014 #6

    Fredrik

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    When i>j, we have ##|a_i-a_j|=a_i-a_j##. Do you know a number that's greater than ##a_i##? If so, you can continue the calculation. I'll denote the number that's known to be greater than ##a_i## by X. We have ##|a_i-a_j|=a_i-a_j<X-a_j##. Now think about doing something similar to ##a_j##. Is there a number Y such that ##X-a_j<X-Y##?
     
  8. Apr 29, 2014 #7

    Fredrik

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    By the way, you can't subtract inequalities. 4<5 minus 1<3 would be 3<2. If you want to use them both, you have to do something like this: 4 < 5 = 4+1 < 4+3.

    That last step follows from one of the axioms for inequalities that are part of the definition of the real numbers:

    1. If x,y>0, then xy>0.
    2. If x<y, then x+z<y+z.
     
  9. May 1, 2014 #8
    After consulting my professor this is the route that he intended for us to take (since we had not discussed the least upper bound).

    It was a pretty long proof written out formally but I'm happy to know how to approach this problem now. Thanks for the help everyone!
     
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