u_{n+1} = 2u_n + (2n-1) squaredu_{n-1},
whence
u_{n+1} - (2n + 1)u_n = -(2n - 1){u_n - (2n - 1)u_{n-1}},
and we readily find that
p_n 1 1 1 1
----- = 1 - -- + -- - -- + ... +- ------,
q_n 3 5 7 2n + 1
whence the value of the fraction taken to infinity is 1/4[pi].
It is always possible to find the value of the n^{th} convergent to a
recurring continued fraction. If r be the number of quotients in the
recurring cycle, we can by writing down the relations connecting the
successive p's and q's obtain a linear relation connecting
p_{nr+m}, p_{(n-1)r+m}, p_{(n-2)r+m},
in which the coefficients are all constants. Or we may proceed as
follows. (We need not consider a fraction with a non-recurring part).
Let the fraction be
a1 a2 a_r a1
-- -- --- --
b1 + b2 + ... + b_r + b1 + ...
p_{nr+m} a1 a2 a_r
Let u_n = --------; then u_n = -- -- ------------, leading
q_{nr+m} b1 + b2 + ... + b_r + u_{n1}
to an equation of the form Au_{n}u_{n-1} + Bu_n + Cu_{n-1} + D = 0,
where A, B, C, D are independent of n, which is readily solved.
2. _The Convergence of Infinite Continued Fractions._--We have seen that
the simple infinite continued fraction converges. The infinite general
continued fraction of the first class cannot diverge for its value lies
between that of its first two convergents. It may, however, oscillate.
We have the relation p_{n}q_{n-1} - p_{n-1}q_n = (-1)^{n}b2b3...b_n,
p_n p_{n-1} b2b3 ... b_n
from which --- - ------- = (-1)^n ------------, and the limit of the
q_n q_{n-1} q_{n}q_{n-1}
right-hand side is not necessarily zero.
The tests for convergency are as follows: