Yeah, I tried the same thing
Ahh well. Thanks for the help though, those marks are long lost.
Computer Science Canada Limit of a series |
| Author: | Martin [ Tue Oct 05, 2004 11:18 pm ] |
| Post subject: | Limit of a series |
Alright, I know how to find the limit of a sequence. A series, on the other hand, is providing me with an endless amount of trouble. Here is my sequence: a(1) = 1, a(n+1) = 1 + 1/a(n) Apparently it converges (has a sum). Also, apparently the limit of this sequence is root 2. Now, can someone please explain to me how to do this? |
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| Author: | AsianSensation [ Fri Oct 15, 2004 2:53 pm ] |
| Post subject: | |
probably too late, but anyways. a(n+1) = 1 + 1/a(n) let Lim a(n+1), as n gets larger and larger be k. take Limit of both sides. Lim a(n+1) = Lim (1 + 1/a(n)) which becomes K = 1 + 1/K K^2 = K + 1 K^2 - K - 1 = 0 K = (1 + or - sqrt (1 + 4))/2 K = (1 + sqrt (5))/2 So, yeah, you sure it's root 2? I get Golden Ratio....... |
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| Author: | Andy [ Sat Oct 16, 2004 1:36 pm ] |
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azn, u solvoed for Lim a(n+1), he wanted to know the limit of the series.. not the sequence |
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| Author: | AsianSensation [ Sat Oct 16, 2004 9:11 pm ] |
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oh, series, my bad. then I have not teh learn3d teh leet math way. |
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| Author: | Martin [ Sun Oct 17, 2004 1:11 am ] |
| Post subject: | |
Yeah, I tried the same thing
Ahh well. Thanks for the help though, those marks are long lost. |
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