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I've been working on an extremely complicated infinite sequence for the past week, and long story short I need to prove that an infinite series found in the sequence converges to one.
I have the first 21 terms of the series, and it appears to loosely resemble a 0.5*ln(x) graph, however about every iteration seems to increase then decrease by more.
I need to find an expression to predict all future terms so I can prove the series converges. Pic related.
Not shown are (1,.5) and (2,.25)
First and second shown values are (3,.0625) and (4,.0625) only the 3rd and 4th terms are known to be equal.
How can I go about solving this? The difference between terms seems to be random, yet appears to be approaching 0 (it's not a simple a+bx+cx^(2)+. . .).
I have found the denominator expression of the series. I'm just stuck on the numerator. (It is likely an alternating power and/or exponential sequence, or maybe with a sinosudial part)
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The first 4 points look like outliers, I wouldn't consider them at first.
Well, I have no idea, nigga. Try to interpolate all the points on the graph using a numerical method, and see what it gives you.
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>>8893708
Thanks, I believe you're correct, the first 3 terms are a bit strange.
I've found a couple graphs with my calculators regression functions (0.803x^-2.1314=y (in blue) and .1411×.773^(x)=y (in purple)) that resembles it, but I need it to be exact
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If it's an alternating series, try splitting it into two subsequences of even and odd terms so it becomes monotone, and then taking the difference.
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>>8893685
The denominator equals 2^([log2(3^n)]+1)
The numerator is 1, 1, 1, 2, 3, 7, 12, 30, 85, 173, 476, 961, 2652, 8045, 17637,...
For n>=0 and where [ ] is an integer function (truncate the decimal)
>>8893747
Thanks for the suggestion, I've tried that but the alternating part is a bit rugged. See the numerator terms. Each one is either roughly a half or a third of the next value, but every so often the alternating part restarts or something.
See 476*2=~961*3=~2652*3=~8045 where it usually alternates 2 then 3 then 2. I've found out that it's because it's dependant on the denominator.
Also (so far) every difference between terms (even every other term) is unique. It's the same way when I applied a few variations of a logarithmic expression to find the numerator expression.
The only clue I have for the numerator is that 0 <numerator<denominator
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Can you explain a bit more about the numerator? If I saw correctly on google, it has to do with the 3n+1 conjecture (which I assume is about 3n+1 and n/2 becoming 1)
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>>8893991
Also: With your sequence equaling 1 you want to show that the denominator is a perfect approximation of the numerator sequence for large n, right?
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The series needs to converge to 1 for any n value, so this is not an issue of numerator/denominator = 1 for large n but that every series will terminate at 1 (or short cycle repeat 4-2-1).
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Yes, this is regarding 3n+1 and 2n...
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>>8894793
Where n is the place of the term
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