Prove Cauchy

**tsang** · August 23rd 2010, 02:27 AM

Hi, I cannot do this question, can anyone please help me?

Let (an) be the sequence given by a0=0 and an=(n+1)/n for all n greater or equal to 1. Prove directly from the definition that (an) is Cauchy.

Cauchy definition should be the one that talks about absolute value of difference between two terms is less then epsilon, right? I tried for a long time, but still don't know how to do it. Can anyone please help me?

**Ares_D1** · August 23rd 2010, 03:08 AM

Start by showing that the sequence converges. (Hint: Rewrite the sequence as $(a_n) = \frac{(n+1)}{n} = 1 + \frac{1}{n}$ to get an idea of what it might converge to, and let N be the smallest natural number satisfying $N > \frac{1}{\epsilon}$ ).

Once you've established that it converges, you can then prove it is Cauchy directly from the definition. Follow the methodology used in Theorem 31 in the lecture notes.

**tsang** · August 23rd 2010, 03:18 AM

Originally Posted by Ares_D1

Start by showing that the sequence converges. (Hint: Rewrite the sequence as $(a_n) = \frac{(n+1)}{n} = 1 + \frac{1}{n}$ to get an idea of what it might converge to, and let N be the smallest natural number satisfying $N > \frac{1}{\epsilon}$ ).

Once you've established that it converges, you can then prove it is Cauchy directly from the definition. Follow the methodology used in Theorem 31 in the lecture notes.

Thanks for your reply.
I actually used this way before, and straight proved sequence is converges.
However, I read the question again which says "use DIRECTLY from DEFINITION", which made me reject my proof. This is not really from definition, right? Definition really refers to the one with $(a_m)-(a_n)}$ , right? That is really the one called "definition of Cauchy". So I'm not sure if I should use this normal way.

**Ares_D1** · August 23rd 2010, 03:38 AM

Originally Posted by tsang

Thanks for your reply.
I actually used this way before, and straight proved sequence is converges.
However, I read the question again which says "use DIRECTLY from DEFINITION", which made me reject my proof. This is not really from definition, right? Definition really refers to the one with $(a_m)-(a_n)}$ , right? That is really the one called "definition of Cauchy". So I'm not sure if I should use this normal way.

I know what you mean. I think what they're trying to emphasize is that you can't just prove convergence and then simply state "Since the sequence is convergent, it necessarily follows (from Theorem 31) that the sequence is Cauchy." This is a perfectly valid statement, of course, but obviously they want us to take it a step further. So, given that we know the sequence is convergent, we can use this to show that for any $\epsilon > 0$ , there exists some $N \in \mathbb{N}$ such that for all $m, n \geq N$ , we have
$|a_m - a_n| < \epsilon$

Which is the definition of a Cauchy sequence. Again, this is essentially identical to the method used in Theorem 31, adapted to this particular sequence.

**HallsofIvy** · August 23rd 2010, 03:43 AM

Originally Posted by tsang

Hi, I cannot do this question, can anyone please help me?

Let (an) be the sequence given by a0=0 and an=(n+1)/n for all n greater or equal to 1. Prove directly from the definition that (an) is Cauchy.

Cauchy definition should be the one that talks about absolute value of difference between two terms is less then epsilon, right? I tried for a long time, but still don't know how to do it. Can anyone please help me?

For two integers, m and n, $a_m- a_n= \frac{m+1}{m}- \frac{n+1}{n}$ $= \frac{n(m+1)- m(n+1)}{mn}= \frac{n+ m}{mn}$ .

If n> m, then $\frac{n+ m}{mn}< \frac{2n}{mn}= \frac{2}{m}$
If m> n, of course, $\frac{n+ m}{mn}< \frac{2m}{mn}= \frac{2}{m}$

Now, what happens as m and n both go to infinity?

**HallsofIvy** · August 23rd 2010, 03:44 AM

Originally Posted by Ares_D1

Start by showing that the sequence converges. (Hint: Rewrite the sequence as $(a_n) = \frac{(n+1)}{n} = 1 + \frac{1}{n}$ to get an idea of what it might converge to, and let N be the smallest natural number satisfying $N > \frac{1}{\epsilon}$ ).

Once you've established that it converges, you can then prove it is Cauchy directly from the definition. Follow the methodology used in Theorem 31 in the lecture notes.

??? What lecture notes are you referring to?

**tsang** · August 23rd 2010, 03:45 AM

Originally Posted by Ares_D1

I know what you mean. I think what they're trying to emphasize is that you can't just prove convergence and then simply state "Since the sequence is convergent, it necessarily follows (from Theorem 31) that the sequence is Cauchy." This is a perfectly valid statement, of course, but obviously they want us to take it a step further. So, given that we know the sequence is convergent, we can use this to show that for any $\epsilon > 0$ , there exists some $N \in \mathbb{N}$ such that for all $m, n \geq N$ , we have
$|a_m - a_n| < \epsilon$

Which is the definition of a Cauchy sequence. Again, this is essentially identical to the method used in Theorem 31, adapted to this particular sequence.

Thanks a lot. That sounds quite nice. Many times I actually proved the question, but I just always feel it is not right, and then I reject myself.
Thank you for your time.
By the way, how do you know the Theorem 31 from the lecture notes?

**Ares_D1** · August 23rd 2010, 03:58 AM

Originally Posted by tsang

By the way, how do you know the Theorem 31 from the lecture notes?

Because I'm in the same Real Analysis class as you?

Originally Posted by HallsofIvy

??? What lecture notes are you referring to?

Lecture notes from our Real Analysis course. Each week we're given practice "tutorial" sheets that help to elucidate and expand on some of the theorems presented in lectures. This is one such example.

**tsang** · August 23rd 2010, 04:18 AM

Originally Posted by Ares_D1

Because I'm in the same Real Analysis class as you?

Lecture notes from our Real Analysis course. Each week we're given practice "tutorial" sheets that help to elucidate and expand on some of the theorems presented in lectures. This is one such example.

Haha, I wonder. Then, we should meet up tomorrow! I'm actually student rep for our class, therefore, I will speak to lecturer tomorrow after lecture, you might should come down so we can catch up!
You are quite good with Letex Code, I just still don't have time to practice. Keep forgetting.
Then, in this case, how about Problem 11 and Problem 16? They both like I think I got them, but I keep rejecting myself because I don't trust myself.

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