Eagle eyed. Edited.(Original post by Felix Felicis)
Pardon me if I'm wrong, but should the second term of your second bracket be ?
Problem 353*/**
For a positive integer , let be the total sum of the intervals of such that in .
Find .
x
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Last edited by Pterodactyl; 23092013 at 19:43. 
bogstandardname
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Using the same method as LOTF's very pretty solution.
I did the final integration by noting that then splitting using partial fractions.Last edited by bogstandardname; 23092013 at 22:22. 
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 24092013 10:26
Solution 351
Using sigma notation, putting the nth terms in to more familiar forms and then letting , we get:
We may evaluate the integral easily using the method of reduction. Alternatively the integral can be obtained instantly through using Ramanujan's master theorem (where ). This gives:
Edit: Just realised this problem is the same as Problem 105 (though my solution is, dare I say, simpler).Last edited by Jkn; 01102013 at 09:01. 
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 24092013 14:31
(Original post by james22)
This is harder than I first thought. I have now failed with 3 methods, all failing because the reccurance relation has an odd number of terms (I think they would work with all ones with an even number of terms), the sums just don't all cancel as 1 is always left out.
I do have a method that will certainly work, but it involves solving teh recurrance relation to get it into an explict form, squaring that and then summing all the geometric series. I don't want to do this ans I doubt it was the intention. 
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 24092013 14:38
(Original post by james22)
I will use the identity
There are several ways to prove this, most straight forward is to use the explicit formula for the fibonacci numbers and it is just an exercise in summing geometric series.
I've started wrestling with that tribonacci question. I agree that the obvious yet ugly technique can yield a solution pretty quickly, though obviously that isn't very satisfying. I've been trying to find some sort of geometric method (i.e. of arranging the squares) but as of yet am having no luck. Have you been trying to do it geometrically or do you think that's a lost cause? 
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 24092013 15:13
Here is one way to do problem 350.
Solution 350
I shall use the following identities which are easily proved by induction (if time permits, I can add their proofs).
We have
;
;
;
.
Now it follows that which is , where is the sum in question.
Spoiler:ShowThis solution is motivated by the Fibonacci case.
We can generalize by replacing the square with an arbitrary positive integer; however, it would not be as easy.
Problem 355*
Let be a set of positive integers. We say that has the property if no element of is a multiple of another. Find the largest subset of having the property .
Problem 356*
In a town there are people and clubs  . If and are even for all , then .Last edited by Mladenov; 24092013 at 16:09. 
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 24092013 16:35
(Original post by Felix Felicis)
Pardon me if I'm wrong, but should the second term of your second bracket be ?
Problem 353*/**
For a positive integer , let be the total sum of the intervals of such that in .
Find .
Problem 354*/**
Evaluate .
Spoiler:Show
First of all, let:
Consider :
Now, just work with the integral:
So:
Now, we can take the limit as n tends to infinity by summing this to infinity:
Summing from n=1 to infinity, we get:
F represents the limit, and we have 2F as after summing we have F(n+2) and F(n+1) remaining, which are equal in the limit. We can rearrange this sum into the form:
Working out F(1) (1) and F(2) (7/6) and adding these on and dividing by two, our final limit is ln4.
Last edited by DJMayes; 24092013 at 20:04. 
Felix Felicis
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 24092013 16:57

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 24092013 17:09
(Original post by Felix Felicis)
Sorry I have to go soon so I don't have time to pinpoint an error  your limit is incorrect but you're along the right lines. The limit should be: 
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 24092013 18:49
(Original post by DJMayes)
Are you absolutely certain of this? I did make a few slips in the evaluation but my final limit is log2.

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 24092013 20:09
(See here: http://en.wikipedia.org/wiki/Riemann_series_theorem ) 
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 24092013 20:31
(Original post by DJMayes)
I have now redone that part of the summation and obtained ln4 as a final answer. However, I'm still not entirely convinced that I made an error; I think infinity (and the fact that the harmonic series is not absolutely convergent) is screwing with me.
(See here: http://en.wikipedia.org/wiki/Riemann_series_theorem ) 
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 24092013 20:40
(Original post by Felix Felicis)
Doh, you're right. When I looked at that last bit of your initial solution, I just rearranged it in a way to get but after looking at it more closely, I see how you got , it wasn't an error on your part, my apologies.
(Or what would be even more interesting is if the limit being the sum of an infinite series that is not absolutely convergent implied that a limit did not exist. However this is entirely speculative and probably more than a bit above my current pay grade. ) 
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 24092013 22:33
(Original post by DJMayes)
Don't worry about it; this was an interesting excursion into one particular paradox of infinity. However I hope there is a way to evaluate this limit without summing infinite series like this as otherwise there is no way to tell which result is the correct limit, or indeed if there is a limit at all.
(Or what would be even more interesting is if the limit being the sum of an infinite series that is not absolutely convergent implied that a limit did not exist. However this is entirely speculative and probably more than a bit above my current pay grade. )
Sorry, I don't get what you mean on your last point, do you mean that if a limit and is not absolutely convergent then the limit does not exist? Well isn't absolutely convergent but the limit still exists.Last edited by Felix Felicis; 24092013 at 22:38. 
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 24092013 22:48
(Original post by Felix Felicis)
Indeed, well my method was more or less the same as your's (I did F(n+1)  F(n) instead) so I've got nada as an alternative to the infinite series. I'll try and hunt the question down again (I found it on AoPS so there'll likely be a solution floating around somewhere, maybe one that doesn't involve the infinite series). Or maybe wait until someone who knows more about analysis shows up.
Sorry, I don't get what you mean on your last point, do you mean that if a limit and is not absolutely convergent then the limit does not exist? Well isn't absolutely convergent but the limit still exists. 
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 25092013 01:07
(Original post by DJMayes)
The last point was completely baseless speculation  if your limit can take any value, do you really have a limit? I do not know and my knowledge of infinite series isn't enough to give any kind of an answer. Either way I am not convinced our method is fine until we've justified which way we sum the series.Last edited by Felix Felicis; 25092013 at 03:12. 
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 25092013 12:25
(Original post by Jkn)
Gonna have to disagree with you there! Not only does that require having the explicit formula in the first place, but it takes a while. Consider the area of the Fibonacci rectangle (it has sides of length and and yet it made up of the squares of the series of Fibonacci numbers)  beautiful! Alternatively a proof by induction follows immediately from Cassini's identity.
I've started wrestling with that tribonacci question. I agree that the obvious yet ugly technique can yield a solution pretty quickly, though obviously that isn't very satisfying. I've been trying to find some sort of geometric method (i.e. of arranging the squares) but as of yet am having no luck. Have you been trying to do it geometrically or do you think that's a lost cause?
I never even considered a geometric method for the tribonacci (I didn't even think about the geometric properties of teh fibonacci numbers either). 
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 25092013 14:41

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 25092013 16:17
Last edited by Felix Felicis; 25092013 at 16:32. 
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(Original post by Mladenov)
Problem 355*
Let be a set of positive integers. We say that has the property if no element of is a multiple of another. Find the largest subset of having the property .x]
The subset . By the pigeonhole principle, any larger subset will have the property .Last edited by MW24595; 25092013 at 17:17.
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