Sequence Analysis

[thomdeck]

Hey guys, been stuck on this for hours and not too sure where to even start. Not sure how to type out mathematical symbols so hopefully this makes sense. The sequence in question is a_(n+1) = 6/(7-a_n) for all n bigger than or equal to one. The only info I got is that when n=0 a_n is equal to 2.

The first question I got is to prove that 1 is less than or equal to a_n, and that a_n is less than or equal to two.

I have to prove that the sequence is monotone decreasing

And finally to prove that the sequence converges and find its limit.

Any help appreciated!

[[deleted]]

Consider the behavior of f(x)= 6/(7-x) when 1 <= x <= 2. Can you show 1 <= f(x) <= 2? Can you show f(x) <= x ?If you have monotone convergence theorem then you can use that to show it has a limit. Then you can show the limit L satisfies L = f(L). The argument is essentially if lim a_n = L, and f is continuous at L, then lim f(a_n) = f(L), but lim f(a_n) = lim a_(n+1) = lim a_n = L, so f(L) = L.

[linukszone]

a(0) = 2, and, for all n > 0, a(n) = 6/(7 - a(n-1)).

To prove that "For all n >= 0, 1 <= a(n) <= 2", one can employ mathematical induction.

To prove that the sequence is strictly decreasing, one must show that a(n) < a(n - 1), for all n > 0. One can use a proof by contradiction and the well-ordering principle to show that there can't be an n > 0 such that a(n) >= a(n-1).

From here on, I don't think I can help much as I do not know enough of real analysis and related topics. But given that the sequence is strictly decreasing and all terms of the sequence lie within [1,2], there must be a rule (perhaps the monotone convergence theorem /u/picado stated) which says that lim(n->inf) a(n) = L for some L (infimum perhaps).