# limits, geometric series, cauchy, proof HELP

by chrisduluk
Tags: cauchy, geometric, limits, proof, series
 P: 56 and can you help me word this one out? Part 1(b): Find conditions on a and r such that the sequence of partial sums of $\sum_{k=0}^\infty ar^k$ is a Cauchy sequence if and only if those conditions are satisfied. (Note that this means that the series is convergent if and only if those conditions are satisfied).
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 Quote by chrisduluk is this ok for my N?
You have to keep in mind that ε may not be an integer (and $\ln 10$ certainly isn't), and N must be an integer. None of the Ns you have suggested are integers.

Don't forget what it is you're trying to do. You're looking for an N such that the following two statements are true:

1. N is a non-negative integer.
2. For all integers n such that n≥N, $\frac{99}{71}(0.01)^{n+1}<\varepsilon$

There is obviously more than one such N.

 Quote by chrisduluk and can you help me word this one out? Part 1(b): Find conditions on a and r such that the sequence of partial sums of $\sum_{k=0}^\infty ar^k$ is a Cauchy sequence if and only if those conditions are satisfied. (Note that this means that the series is convergent if and only if those conditions are satisfied).
I think this is the hardest part of the problem. I will take a look at it.
 P: 56 can you PLEASE help me figure out what my N is supposed to look like? I simply don't know how else to do it other than the way i did it above. I BEG of you. BEG! I need to finish this!!!
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 Quote by chrisduluk can you PLEASE help me figure out what my N is supposed to look like? I simply don't know how else to do it other than the way i did it above. I BEG of you. BEG! I need to finish this!!!
Fredrik is NOT going to give you the answer. You will have to figure it out on your own. We can only guide you to the solution.
 P: 56 then help guide me into finding an N that works! Should it have logs in it? Should the N have epsilon in it? Why was my N wrong above?
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I understand that this is important to you, but I don't want to tell you so much that I'm breaking the forum rules. Let me ask you this, is it at least clear to you that an N with the desired properties exist? Can you explain why such an N must exist?

Problem I (b) is easier than I thought it would be, but it's still kind of hard. You have made it harder for me to explain it to you by not answering the questions I asked here:
 Quote by Fredrik We need to make sure that you understand a few other things: 1. Do you understand what it means to say that a series is convergent? Specifically, if I write $\sum_{k=0}^\infty a_k=s$, do you know what that means? 2. Do you understand what it means to say that a sequence is convergent? Specifically, if I say that $s_n\to s$, do you know that means? 3. Do you understand the definition of Cauchy sequence? Specifically, if I say that $\langle s_n\rangle_{n=0}^\infty$ is a Cauchy sequence, do you know what that means? (That's the notation I use for the sequence $s_0,s_1,\dots$. Your teacher may use something different).
The work you showed me on problem II (b) suggests that you know the answer to questions 1 and 2. Can you at least answer question 3?
 P: 56 i do realize there's an N so it satisfies both of those properties, i just don't know how to eliminate the logs so N is an integer. I've never done an example this ridiculous... All of my examples worked out nicely, so it's clear we'd just make N an integer/epsilon. How do i eliminate the logs when finding my N? Is it simply 71/99E +anything? and i do understand what makes it cauchy, we pick any positive epsilon, and pick any two values down the line of the sequence and say the difference between the two is whatever we want it to be...
 PF Patron Sci Advisor Emeritus P: 8,837 How do you find the N when you prove that 1/n→0? Let ε>0 be arbitrary. We want to prove that there's a natural number N such that for all integers n, n≥N implies |1/n-0|<ε. If we solve this for n, we get n>1/ε. But 1/ε isn't an integer either.
 P: 56 ...so what's that mean? huh? Can you just tell me what part of the N i calculated above is wrong? And how i can make it right? ie remove the denominator, add 1 to it, etc??
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 Quote by chrisduluk and i do understand what makes it cauchy, we pick any positive epsilon, and pick any two values down the line of the sequence and say the difference between the two is whatever we want it to be...
You're going to have to be much more specific about what a Cauchy sequence is when you start working on I (b). It seems to me that you're supposed to use the following:

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.
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 Quote by Fredrik How do you find the N when you prove that 1/n→0? Let ε>0 be arbitrary. We want to prove that there's a natural number N such that for all integers n, n≥N implies |1/n-0|<ε. If we solve this for n, we get n>1/ε. But 1/ε isn't an integer either.
 Quote by chrisduluk ...so what's that mean? huh?
You tell me. This is the simplest possible problem of the same sort that you need to solve. So you should probably put your problem aside for a while, and figure out the answer to this one first.
 P: 56 Fredrik, i'm sorry but i'm not following you and i don't have any more time to work on this. I HAVE AN EXAM TOMORROW MORNING. I haven't even started studying for it yet because i keep running around aimlessly on this! I specifically posted on here to get HELP and i'm just getting more confused. Can you PLEASE try to make things more clear to guide me step by step so i can do these problems? Right now i need to know why my N was wrong. Can you follow my work that i scanned above?
 P: 56 why can't i make my N= [ln(71/99E) / ln(100)] +1
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 Quote by chrisduluk Fredrik, i'm sorry but i'm not following you and i don't have any more time to work on this. I HAVE AN EXAM TOMORROW MORNING. I haven't even started studying for it yet because i keep running around aimlessly on this! I specifically posted on here to get HELP and i'm just getting more confused.
I'm not going to break any forum rules because you have an exam tomorrow. I have given you much more information than you would have needed if you had done a few more exercises before you came here. And I have spent a lot of time giving you that information. So don't act like you haven't gotten any help. We have made significant progress, but we would have made more if you hadn't been so unwilling to write down definitions and answer questions.

 Quote by chrisduluk Can you PLEASE try to make things more clear to guide me step by step so i can do these problems?
I have done that for all parts except what you need to finish I(b), but since you haven't even followed my instructions on how to start I(b) yet, you have no need to for the rest of it yet.

 Quote by chrisduluk Right now i need to know why my N was wrong. Can you follow my work that i scanned above?
What N are you talking about? The one where you say that N is equal to an inequality? That doesn't even make sense.
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 Quote by chrisduluk why can't i make my N= [ln(71/99E) / ln(100)] +1
Do the brackets mean something different than parentheses here?
 P: 56 on the scanned sheet, ignore the inequality, so N= everything to the right of the inequality sign. and you mean the floor function?
 P: 56 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?
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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.