Sequences-- Is a_(n+1) = 1 / (3 - a_n) convergent?

[krisaldine]

Can someone show me how to approach this problem please?

NOTE: sub= subscript

Show that the sequence defined by

(a sub 1) = 2

(a sub (n+1)) = 1/ (3 - (a sub n))

satisfies 0 < (a sub n) is less than or equal to 2 and is decreasing. Deduce that the sequence is convergent and find its limit.

Thank you!
Krisaldine

 

[galactus]

I see a relation to the Fibonacci numbers. The numerators and denominators of the sequence are every other Fibonacci number.

1,1,2,3,5,8,13,21,34,55,89,233,.......


You have in the sequence: 1, 1/2, 2/5, 5/13, 13/34, 34/89, 89/233, 233/610,......


\(\displaystyle \L\\\frac{F_{2n+1}}{F_{2n+3}}\) is what it appears to be.

As you may know, the limit of the ratio of the Fibonacci numbers tends to the Golden Ratio.

So what we would have as a limit is:

\(\displaystyle \L\\\frac{\frac{\sqrt{5}-1}{2}}{\frac{\sqrt{5}+1}{2}}=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\approx{0.38196601125.......}\)

At least. that's what it looks like to me.

 

[daon]

given: \(\displaystyle 0 \,\, < \,\, a_{n} \,\, \le \,\, 2\) for all n and

\(\displaystyle \L

a_{n+1} \,\, = \,\, \frac{1}{3-a_n} \\ ................\)

We already know that \(\displaystyle a_{n} \,\, \g \,\, 0 \,\, \Rightarrow \,\, \frac{1}{3-a_n} \,\, \g \,\, \frac{1}{3-0} \,\, = \,\, \frac{1}{3}\)

So as \(\displaystyle n \rightarrow \infty \,\,\) all \(\displaystyle a_n\) are greater than or equal to 1/3 (that is to say: bounded below by 1/3).

If non-increasing is a GIVEN, then you can claim it has a limit since it is bounded below. If \(\displaystyle a_n\) has a limit then \(\displaystyle lim(a_n) \,\, = \,\, lim(a_{n+1})\). This is a theorem, and hopefully it has been discussed in your class. And hence:

\(\displaystyle lim \,\, a_n \,\, = \,\, lim \,\, a_{n+1} \,\, = \,\, lim \,\, \frac{1}{3- a_{n+1}}\)

By letting \(\displaystyle lim \,\, a_n \,\, = \,\, L\), A.O.L. we get:

\(\displaystyle \L L^2 - 3L +1 =0 \,\, \,\, \Rightarrow \,\, \,\, L= \frac{3 \,\, +/- \,\, \sqrt{5}}{2}\).

Only one of these makes sense.

(Galactus beat me to it )

 

[galactus]

No, we just explained it differently. I like your method. I just mentioned the Fibonacci thing and seen it that way.

The way you explained it is exactly how the Golden Ratio and the Fibonacci relationship is shown.

If we start with the recursive relation: \(\displaystyle f_{n+2}=f_{n+1}+f_{n}\)

and divide through by \(\displaystyle f_{n+1}\), we get: \(\displaystyle r_{n+1}=1+\frac{1}{r_{n}}\)

Let n tend to infinity: \(\displaystyle L=1+\frac{1}{L}\)

And we have the roots of \(\displaystyle L^{2}-L-1\) as the Golden Ratio.

See the relationship?. Cool problem.

 

[daon]

No, we just explained it differently. I like your method. I just mentioned the Fibonacci thing and seen it that way.

The Fibonacci thing is cool. Never would have seen that.

 

[pka]

We are given a sequence defined inductively:
\(\displaystyle a_1 = 2,\quad n > 1\quad a_n = \frac{1}{{3 - a_{n - 1} }}\).

So use induction to prove that it is decreasing bounded.
\(\displaystyle \L 3 > a_1 > a_2 > 0\).

Now suppose that it is true for K.
\(\displaystyle \L \begin{array}{l}
3 > a_K > a_{K + 1} > 0 \\
- 3 < - a_K < - a_{K + 1} < 0 \\
0 < 3 - a_K < 3 - a_{K + 1} < 3 \\
\frac{1}{3} < \frac{1}{{3 - a_{K + 1} }} < \frac{1}{{3 - a_K }} \\
\frac{1}{3} < a_{k + 2} < a_{k + 1} \\
\end{array}.\)

This proves that the sequence is decreasing and bounded below.
This means that it must have a limit.

Say that limit is L. To find the value of the limit solve for L:
\(\displaystyle \L L = \frac{1}{{3 - L}}.\)

 

[krisaldine]

wonderful! Thank you all for your help! I am really starting to enjoy this class!

Thanks again!

Krisaldine