Limits, geometric series, cauchy, proof HELP

[chrisduluk]

can anybody tell me where my cauchy proof doesn't look right so i can fix it??

And how to i answer the "determine the conditions on a and r..." question?

 

[Fredrik]

on the scanned sheet, ignore the inequality, so N= everything to the right of the inequality sign.

and you mean the floor function?

Yes, I was wondering if you meant that [x] denotes the integer part of x (i.e. the largest integer n such that n≤x). (I actually wasn't familiar with the term "floor function"). I highly recommend that you use that function here. How else would you make N an integer?

I'm not sure I understand what you're doing in post #54. You wrote down $71\big(1/100\big)^n < \varepsilon$. But we started with $\frac{71}{99}\big(\frac{1}{100}\big)^{n+1} <\varepsilon$. Is there a first step that you didn't write out, where you replaced 99 by 100? If so, I think you did it wrong. However, there's no need to make any simplifications like replace 99 by 100. Your N doesn't have to be pretty.

 

[Fredrik]

can anybody tell me where my cauchy proof doesn't look right so i can fix it??

And how to i answer the "determine the conditions on a and r..." question?

What do you mean? You haven't even started yet. At least you haven't posted anything about it. Post #64 is telling you how to start.

 

[chrisduluk]

yeah that was an error... if you only knew how much crap i have scattered around me right now... so is this how i write the proof? if not, can you point out or fix the errors?

 

[Fredrik]

yeah that was an error... if you only knew how much crap i have scattered around me right now... so is this how i write the proof? if not, can you point out or fix the errors?

The first inequality under the horizontal line is unnecessary but correct. The second is correct too when n≥N, but you don't seem to have considered these inequalities together. You're creating a string of inequalities that look like this ε > something < something < f(N). And then you appear to be solving f(N)>ε for N. But your string of inequalities did nothing to prove that f(N)>ε.

You have convinced me that you understand the general idea now (even though you still seem to have difficulties expressing it), so I will tell you a bit more. When you hand it in, you don't have to explain how you found your N. You just have to prove that your choice of N works, i.e. that it ensures that for all integers n such that n≥N, we have
$$\frac{71}{99} \left(\frac{1}{100}\right)^{n+1} <\varepsilon.$$ You will have to prove that even if you do explain how you found your N.

If you want to explain how to find N, then do it like this: We're looking for a natural number N such that the inequality above holds for all n≥N. If the inequality holds for all n≥N, it holds for n=N. So now you can just set n=N in the inequality and solve for N. You get a result of the form N>f(ε). This tells you that f(ε) is a lower bound on the set of acceptable choices. This obviously means that you can choose N to be any integer that's >f(ε). It's simple and convenient to choose the smallest one: N=[f(ε)]+1.

When you first solved for n in post #50, you found your f(ε) (the right-hand side of the last inequality). All you had to do after that was to choose N=[f(ε)]+1.

 

[chrisduluk]

but in post 50 i had 9900 as the denominator... should it be 9900 or 99?

 

[Fredrik]

It's 99 if the exponent is n+1, 9900 if the exponent is n (as in post #50).

 

[chrisduluk]

ok, so how would the "string of inequalities" look when i enter in big N? In all of the examples I've done, the string ended with something like

if N=90/361E...

90/361n < 90/360N < 90/360(90/361E) = E
but in this problem I'm not getting the last thing to be =E

 

[Fredrik]

I don't understand what you're doing there. As I said in #75, the best way to find N is to realize that the equality holds for n=N. So when you "solve for n", you're really solving for N. You can even set n=N in the inequality before you begin. And as I said earlier, there's no need to simplify anything. So you don't need a string of inequalities. You just need to set n=N in the inequality you started with and solve for N.

 

[chrisduluk]

and once i solve for N, how do i show that abs(Sn - thing it converges to) < E?

 

[Fredrik]

Ah, I see. You do need a string of inequalities for that. I would use that n≥N as early as possible. You have already showed that |s_n-thing it converges to|=(71/99)(1/100)^{n+1}, and now you should immediately use that n≥N. Then you use your choice of N.

Edit: This might be slightly easier if you use base-10 logarithms when you solve for N. You will still have to use some stuff you know about logarithms to work it out to the end. (I just did, so I know it can be done).

 

[chrisduluk]

k, but if i pick my N as blah blah +1 i don't get a nice = E in the end. I get 1/100E. But if i just keep blah blah without the +1 it works out, like below:

 

[Fredrik]

k, but if i pick my N as blah blah +1 i don't get a nice = E in the end. I get 1/100E. But if i just keep blah blah without the +1 it works out, like below:

That's fine, because N=[x]+1 implies N≥x, and that's the inequality you actually use to evaluate (1/100)^n. $n\geq N\geq x$ implies
$$\frac{1}{100^n}\leq\frac{1}{100^N}\leq\frac{1}{100^x}$$

 

[chrisduluk]

so my proof looks good? :D

 

[Fredrik]

Yes. Only I(b) left now. Not sure I will stay up much longer though. I'm thinking about asking someone else to take over. Do you want to continue?

 

[chrisduluk]

if you can, yes please. I'm clueless.

 

[Fredrik]

You will have to start with what I said in post #64. I'm going to bed, but I will make a post in the Science Advisor forum (hidden for normal users) asking if someone else can take over. I have already prepared some instructions for the person who takes over, including the solution to the problem. The good news is that the problem is much easier than I thought at first.

 

[chrisduluk]

oh thank you! It doesn't involve cauchy, right?

 

[Fredrik]

It does. That's the only one of the problems that does involve Cauchy sequences or the Cauchy criterion. I have posted a request for someone to take over now. Not sure how long it will take for someone to show up. It's 4:40 AM in Europe, so a lot of people have already gone to bed.

 

[chrisduluk]

oh damn... yeah, i didn't use cauchy for that. I just did this. But I'm guessing i'll need cauchy.

Wow, you're in europe! That's awesome, where i am in the US it's 10:43pm!

If no one else chimes in, what should i do? Only because i need to get some sleep myself.. This assignment is due 6 hours from now, and that would be the minute before i have class.

Here's what i did already, which I am sure is wrong

 

[chrisduluk]

so do they want something like this? If I'm not using the .7171717, what will the beginning of my proof look like? Will be it to prove (a-ar^n+1)/(1-r) converges to a/(1-r)?

 

[Fredrik]

Do you remember that when you started this thread, you insisted that you would have to use the Cauchy criterion? We haven't had any use for it in I(a), II(a) or II(b), so if you were right from the start, then we have to use it now. I told you what it means to use the Cauchy criterion, and how you should start, in post #64:

Definition: A series is convergent if and only if its sequence of partial sums is convergent. If the sequence is convergent, its limit is called the sum of the series.
Theorem: A series with real terms is convergent if and only if its sequence of partial sums is a Cauchy sequence.

You need to use this theorem to prove that your series is convergent. So the first thing you should write down is exactly what it means for your sequence of partial sums to be a Cauchy sequence.

Since then, I have told you at least twice that you need to read this and do what I'm saying at the end. I guess you're disappointed that no one else showed up, but if someone had, I would have wanted them to tell you nothing until after you have done this first step. They would have been wrong to do anything else.

Note that I told you that this is what it means to use the Cauchy criterion in post #5, 32 hours ago. I said "I assume", but you yourself confirmed that I was right in #6, half an hour later. Posts #9 (by micromass) and #25 (by me) also told you that I(b) is where you have to use the Cauchy criterion. #64 just spelled it out in as much detail as possible without violating forum rules.

One thing I didn't see from the start (this is the sort of thing you don't see until you actually start working on the problem) is that your teacher isn't just forcing you to use the Cauchy criterion so that you will have to learn it. He's doing it because if you just prove that $\sum_{k=0}^\infty ar^k=\frac{a}{1-r}$ when |r|<1, you don't know if the series fails to converge for other values of r, or if the value of a is relevant when |r|≥1.

If you start by writing down what it means for the sequence of partial sums to be a Cauchy sequence, the rest is fairly straightforward. Keep in mind that a sequence of real numbers is convergent if and only if it's a Cauchy sequence. That's all I can tell you until you have actually done that first step.