a_1 = 1, a_n+1 = 1/(1+a_n) for n >= 1. deduce convergence

[dts5044]

6. Let a_1 = 1 and a_n+1 = 1/(1+a_n) for n>=1

a) prove that 0 < a[sub:1an8mau5]2k[/sub:1an8mau5] < a[sub:1an8mau5]2k+2[/sub:1an8mau5] < a[sub:1an8mau5]2k+1[/sub:1an8mau5] < a[sub:1an8mau5]2k-1[/sub:1an8mau5] for integers k >= 1

I solved this, and here's how:

Let a_n = the ratio of two positive real numbers, p and q. (a_n = p/q)
then a_n+1 = q/(q + p)

I then used an induction proof, calculating the first four terms and showing the above inequality is true for k = 1.

Then, assuming 0 < a[sub:1an8mau5]2k[/sub:1an8mau5] < a[sub:1an8mau5]2k+2[/sub:1an8mau5] < a[sub:1an8mau5]2k+1[/sub:1an8mau5] < a[sub:1an8mau5]2k-1[/sub:1an8mau5] for k = n

and letting a[sub:1an8mau5]2k+1[/sub:1an8mau5] = p/q, a[sub:1an8mau5]2k+2[/sub:1an8mau5] = q/(q + p)

we can rewrite a[sub:1an8mau5]2k+2[/sub:1an8mau5] < a[sub:1an8mau5]2k+1[/sub:1an8mau5] as q/(q + p) < p/q
i.e. 0 < p[sup:1an8mau5]2[/sup:1an8mau5] + pq - q[sup:1an8mau5]2[/sup:1an8mau5]

for k = n+1 prove:
0 < a[sub:1an8mau5]2k+2[/sub:1an8mau5] < a[sub:1an8mau5]2k+4[/sub:1an8mau5] < a[sub:1an8mau5]2k+3[/sub:1an8mau5] < a[sub:1an8mau5]2k+1[/sub:1an8mau5]

rewriting in terms of p and q, each inequality can be rearranged and proven using 0 < p[sup:1an8mau5]2[/sup:1an8mau5] + pq - q[sup:1an8mau5]2[/sup:1an8mau5].

b) Deduce the sequence a_n is convergent, and find its limit

this is the part I have trouble with. I think you divide the sequence into two subequences, the terms when n is even and the terms when n is odd. The even subsequence is an increasing function bounded by the odd subsequence, and the odd subsequence is a decreasing function, bounded by the even sequence. I'm not sure if this proves convergence or not. If it does, how do you find the limit? I assumed that the limit of each subsequence is where the odd and even subsequences equal one another, but my professor said this was not necessarily true when I presented the answer. Can anyone help?

also, another problem I had is very similar:
a_1 = 1 and a_n+1 = 1 + 1/(1+a_n)

a) prove that 1 <= a[sub:1an8mau5]2k-1[/sub:1an8mau5] < a[sub:1an8mau5]2k+1[/sub:1an8mau5] < a[sub:1an8mau5]2k+2[/sub:1an8mau5] < a[sub:1an8mau5]2k[/sub:1an8mau5] <= 3/2 for integers k >= 1

I used the same method as in the previous problem to prove this by induction

b) deduce the sequence is convergent, and find its limit

same as last time, I cannot prove convergence or find the limit

can someone help? on one or both! thank you!

 

[pka]

Theorem: Any monotone, bounded sequence converges.
A bounded non-decreasing sequence converges to its least upper bound.
So as a result of parts 1 & 2 we know that this sequence converse.
So say \(\displaystyle \left( {a_n } \right) \to L.\)
Now all you have to do is solve for L: \(\displaystyle L = \frac{1}{{1 + L}}\)

 

[dts5044]

but is it safe to assume that the odd subsequence of a[sub:3a53249u]n[/sub:3a53249u] is the least upper bound of the even subsequence?

In order for the sequence to converge, the two subsequences (odd and even values of n) have to converge to the same limit, correct?

Sorry, but I'm still a little hazy on how you prove convergence. Could you explain how you prove both subsequences converge to the same place?

 

[pka]

but is it safe to assume that the odd subsequence of a[sub:2u529nfm]n[/sub:2u529nfm] is the least upper bound of the even subsequence?

THEOREM: If a sequence converges then any subsequence converges to the same limit.