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).
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).
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