Limits of Monotonic Sequences

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If I have monotonic sequence, would it suffice to analyze |a(n)-a(n-1)| as n gets large? I know for Cauchy sequences, you have to analyze every term after N, but for monotonic sequences that are also Cauchy, can you just analyze the difference between consecutive terms?
 

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  • #2
andrewkirk
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Analyse in order to deduce what? Convergence to a limit? If a sequence is monotonic the easiest way to prove it converges is usually to show that it is bounded above and then use the monotone convergence theorem.
 
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  • #3
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Analyse in order to deduce what? Convergence to a limit? If a sequence is monotonic the easiest way to prove it converges is usually to show that it is bounded above and then use the monotone convergence theorem.
I'm writing a computer program to show convergence. Say the function is f(n)=n/(n+1). Now we know that the limit is 1. But its kinda cheating to use that knowledge in the program since that somewhat defeats to purpose of finding the limit numerically, so I just run the iterations up until the difference between consecutive terms is less than some threshold.

Then I return the iteration number and the current term of the sequence to see how fast it converges and get an approximation of the limit.
 
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I suppose my reasoning is if a continuous function were to approach a limit, has to be concave down(or concave up depending), hence a sequence embedded in that function would have decreasing absolute differences between consecutive terms as n increases without bound.
 
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andrewkirk
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I suppose my reasoning is if a continuous function were to approach a limit, has to be concave down(or concave up depending), hence a sequence embedded in that function would have decreasing absolute differences between consecutive terms as n increases without bound.
They will eventually decrease but, without additional info, we cannot conclude anything from current differences about bounds on future differences. A function may appear to be asymptotically approaching a limit from above but then bump up significantly before finally approaching a limit.

eg consider

$$f(x)=e^{-x} + e^{-\frac{(x-1000)^2}2}$$
which has a Gaussian bump centred at ##x=1000## in what is otherwise an asymptotic, concave-up, slide towards a limit of 0 as ##x\to\infty##.

Even more pathological functions could be created that bump up infinitely many times. One possibility may be:
$$g(x)=
e^{-x} + \sum_{k=1}^\infty e^{-\frac{(x-k\cdot 1000)^2}2}$$
I have not proven that the function is well-defined, although I suspect it is.
 
  • #6
mathwonk
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if the sequence is cauchy, then it is convergent and that does not need to be checked further. if you do not know the sequence is cauchy, then i do not see how to check this computationally; i.e. even if it is monotone, the successive differences do not need to be monotone, and even if they are, the sequence does not need to converge (1, 1+1/2, 1+1/2+1/3, 1+1/2+1/3+1/4,.....). and even if the sequence does converge, and the successive differences are monotone, there is no number small enough, so that once the differences get that small, you know how close you are to the limit. I.e. you could know that at some point all later successive differences are less than 1/1000,000,000,000, and still the limit could be a billion larger than your current estimate.
 

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