
#55
Dec1411, 04:08 PM

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 [itex]\sum_{k=0}^\infty ar^k[/itex] 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). 



#56
Dec1411, 04:18 PM

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PF Gold
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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 nonnegative integer. 2. For all integers n such that n≥N, [itex]\frac{99}{71}(0.01)^{n+1}<\varepsilon[/itex] There is obviously more than one such N. 



#57
Dec1411, 04:29 PM

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!!! 



#58
Dec1411, 04:33 PM

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#59
Dec1411, 04:35 PM

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?




#60
Dec1411, 04:44 PM

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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: 



#61
Dec1411, 04:59 PM

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... 



#62
Dec1411, 05:21 PM

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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/n0<ε. If we solve this for n, we get n>1/ε. But 1/ε isn't an integer either.




#63
Dec1411, 05:25 PM

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?? 



#64
Dec1411, 05:28 PM

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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. 



#65
Dec1411, 05:30 PM

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#66
Dec1411, 05:31 PM

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? 



#67
Dec1411, 05:40 PM

P: 56

why can't i make my N= [ln(71/99E) / ln(100)] +1




#68
Dec1411, 05:43 PM

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#69
Dec1411, 05:47 PM

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#70
Dec1411, 05:50 PM

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? 



#71
Dec1411, 06:05 PM

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? 



#72
Dec1411, 06:15 PM

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I'm not sure I understand what you're doing in post #54. You wrote down [itex]71\big(1/100\big)^n < \varepsilon[/itex]. But we started with [itex]\frac{71}{99}\big(\frac{1}{100}\big)^{n+1} <\varepsilon[/itex]. 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. 


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