# Limits of Monotonic Sequences

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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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.

• FallenApple
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.

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.

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.

mathwonk