# Prove that a monotone increasing and bounded sequence converges

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).

Related Calculus and Beyond Homework Help News on Phys.org
mfb
Mentor
##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##.

Fredrik
Staff Emeritus
Gold Member
$$L - \epsilon < a_{i} ≤ L$$
$$L - \epsilon < a_{j} ≤ L$$
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.

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 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.

##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##.
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?

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'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.

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.
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?

jbunniii
Homework Helper
Gold Member
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.
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##.

Fredrik
Staff Emeritus
Gold Member
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.
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##?

Fredrik
Staff Emeritus
Gold Member
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.

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##.
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!