3. Show convergence of infinite simple continued fractions.

Show the convergence of the infinite simple continued fraction in Problem 2, representing the recurrence relation {a_n=a_{n-1}+a_{n-2}}.

Let {\dfrac{p_ {n}}{q_{n}}} be the fraction represented by the {n}th term in the continued fraction. Thus {\dfrac{p_ {2n}}{q_{2n}}} are the even convergents; {\dfrac{p_ {2n+1}}{q_{2n+1}}} are the odd convergents; both {p_n} and {q_n} satisfy the recurrence relation.

One needs to show that {\dfrac{p_{2n}}{q_{2n}}} is bounded and monotonic increasing and {\dfrac{p_{2n+1}}{q_{2n+1}}} is bounded and monotonic decreasing, and that these two subsequences converge to the same limit.

A formula for the difference of convergents will help:

{\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{p_nq_{n-1}-p_{n-1}{q_n}}{q_nq_{n-1}}}.

Expanding the numerator on the right side by substitution from the recurrence relation,

{p_nq_{n-1}-p_{n-1}q_n=(p_{n-1}+p_{n-2})q_{n-1}-p_{n-1}(q_{n-1}+q_{n-2})=-(p_{n-1}q_{n-2}-p_{n-2}q_{n-1})}.

Repeat the substitution on the right until the result is in terms of the sequence initial constants:

{p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}(p_1q_0-p_0q_1)=(-1)^{n-1}}.

Thus, {\dfrac{p_n}{q_n}-\dfrac{p_{n-1}}{q_{n-1}}=\dfrac{(-1)^{n-1}}{q_nq_{n-1}}}.

Similarly, {\dfrac{p_n}{q_n}-\dfrac{p_{n-2}}{q_{n-2}}=\dfrac{(-1)^{n-1}}{q_nq_{n-2}}}.

Since {q_n>0}, {\dfrac{p_{2n}}{q_{2n}}-\dfrac{p_{2n-2}}{q_{2n-2}}>0} and {\dfrac{p_ {2n}}{q_{2n}}} is monotonic increasing.

Similarly {\dfrac{p_{2n+1}}{q_{2n+1}}-\dfrac{p_{2n-1}}{q_{2n-1}}<0} and {\dfrac{p_ {2n+1}}{q_{2n+1}}} is monotonic decreasing.

Both subsequences are bounded, for {\dfrac{p_ {2n}}{q_{2n}} < 2} and {\dfrac{p_ {2n+1}}{q_{2n+1}}>1}.

Thus it remains to show that both subsequences converge to the same limit, by showing the difference between the sequences tends to {0} for large {n}. Observing that {q_n>n} when {n>3}:

{\left| \dfrac{p_{2n}}{q_{2n}}-\dfrac{p_{2n-1}}{q_{2n-1}} \right|=\dfrac{1}{q_{2n}q_{2n-1}}\le \dfrac{1}{2n(2n-1)}\rightarrow 0}.

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