Continued Fractions, Convergence Theory. Vol. 1, 2nd Editions. Loretzen, Waadeland. Atlantis Press. 2008

12 Chapter 1: Introductory examples
In fact, we can do even better! The tail values
f (n) := 30+0.9n+1
1 +
30+0.9 n+2
1 +
30 + 0.9 n+3
1 +···
satisfy the recurrence
f (n) = 30+0.9n+1
1+f (n+1) , i.e., f (n) (1 + f (n+1) )=30+0.9 n+1 .
Assume that f (n) =5+cr n+1 + o(r n+1 ) for some c ∈ R and |r| < 1, where o(r n+1 )
is the usual “ little o-notation” defined by
u n = o(r n+1 u n
) ⇐⇒ lim =0.
n→∞ rn+1 That is, we assume that c is chosen such that
Since then
f (n) − (5 + cr n+1 )
lim
=0.
n→∞ r n+1
f (n) (1 + f (n+1) ) = (5 + cr n+1 )(6 + cr n+2 )+o(r n+1 )
=30+c(6 + 5r)r n+1 + o(r n+1 ) ,
we must have r =0.9 and c(6+5r) = 1; i.e., c =2/21. Therefore the approximants
S n (w n ) with w n =5+ 2
21 · 0.9n+1 should be an even better choice than S n (5), in
particular for large n. This is in fact true. We shall return to this idea in Section
5.1.5 on page 232. The table below gives a very clear indication that this is so. The
value f is in this case f =5.085066164924 correctly rounded to 12 decimal places.
✸
n S n (0) f − S n (0) f − S n (5) f − S n (w n )
1 30.9000 −25.8 −6.5 · 10 −2 4.4 · 10 −4
2 0.97139 4.1 4.8 · 10 −2 −3.0 · 10 −4
3 15.6770 −10.6 −3.7 · 10 −2 2.0 · 10 −4
4 1.85765 3.2 2.7 · 10 −2 −1.4 · 10 −4
35 5.10127 −1.6 · 10 −2 −3.8 · 10 −6 7.3 · 10 −10
36 5.07160 1.3 · 10 −2 2.8 · 10 −6 −4.9 · 10 −10
37 5.09631 −1.1 · 10 −2 −2.1 · 10 −6 3.3 · 10 −10
38 5.07571 9.3 · 10 −3 1.6 · 10 −6 −2.2 · 10 −10
85 5.08507 −1.8 · 10 −6 −2.1 · 10 −12 2.1 · 10 −18
86 5.08506 1.5 · 10 −6 1.6 · 10 −12 −1.4 · 10 −18
87 5.08507 −1.2 · 10 −6 −1.2 · 10 −12 9.7 · 10 −19
88 5.08507 1.0 · 10 −6 9.0 · 10 −13 −6.5 · 10 −19