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Proof of Convergence of a (recursive) sequenceHi, I'd like to proof, if the following sequence converges; if yes, what's the limes? It's: a_1 > 0 und a_(n+1) = 1/(1+a_n) I thought I should show 1) The sequence is bounded (above or below) 2) The sequence is monotonous.. But I do not know where to begin =/ |

Re: Proof of Convergence of a (recursive) sequenceQuote:
The way you get started on a problem like this is to just plug in numbers to get a feel for the sequence. I took a = 2. Actually this was an accident, for some reason I thought it said a > 1 so I picked 2. But re-reading, it says a > 0. So you should start by trying this for a = 1. a1 = 2 a2 = 1/(2+1) = 1/3 a3 = 1/(3/3 + 1/3) = 1/(4/3) = 3/4 a4 = 1/(4/4 + 3/4) = 1/(7/4) = 4/7 a5 = 1/(7/7 + 4/7) = 1/(11/7) = 7/11 ... The reason I wrote it out like that is that the process of adding 1 to a rational number is: 1 + (n/m) = (m/m) + (n/m) = (n+m)/m or "finding the common denominator" like they say in grade school. So you can begin to see the pattern. The numerators and denominators of your sequence are the terms of 3, 4, 7, 11, ..., but offset by one. I looked up 3, 4, 7, 11, 18, 29, 47, ... in OEIS, and it turns out to be the Lucas numbers (beginning at 2): L(n) = L(n-1) + L(n-2)http://oeis.org/A000032 Now I looked up Lucas numbers on Wikipedia. https://en.wikipedia.org/wiki/Lucas_number Lucas numbers are a special case of Lucas sequences. A Lucas sequence is a sequence in which a_(n+2) = a_n + a_(n+1). So it's actually a generalized Fibonacci sequence. In fact before reading the Wiki page on Lucas numbers, I perused the separate page on Lucas sequences ... https://en.wikipedia.org/wiki/Lucas_sequence In fact the concept of the Lucas sequence has as special cases the Fibonacci numbers and other important number sequences. (All these are integer sequences. Though I don't see why you wouldn't also be interested in letting the terms be real, or even complex. What you really care about is that each term is the sum of the previous two. In any event, now you put in 2, 1 as the first pair of numbers and you get 2, 1, 3, 4, 7, 11, 18, ... and these are the Lucas numbers. I'm going into all this detail because you asked how you get started on a problem like this. And the answer is that you can use traditional means, such as plugging in actual numbers and getting a feel for the situation and starting to see the pattern ... and you can combine that with modern methods such as looking things up on the Internet. Anyway it's pretty clear that for a = 2, your sequence is bounded by 0 and 1, and therefore converges. And I'm sure sure there must be some really cool closed-form expression for the limit. Now you have to see if you can make sense of the general case with a > 0. I'm not saying my approach is the right one, it's probably not. But it's an example of how you can begin to get a handle on the problem. |

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