# Prove that a monotone increasing and bounded sequence converges

1. Apr 29, 2014

### Tsunoyukami

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. Apr 29, 2014

### Staff: 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 $a_i$.

3. Apr 29, 2014

### Fredrik

Staff Emeritus
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.

4. Apr 29, 2014

### Tsunoyukami

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?

5. Apr 29, 2014

### jbunniii

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 $L$.

6. Apr 29, 2014

### Fredrik

Staff Emeritus
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 $X-a_j<X-Y$?

7. Apr 29, 2014

### Fredrik

Staff Emeritus
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.

8. May 1, 2014

### Tsunoyukami

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!