By Peter J. Cameron

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Now the recurrence relation qk = ak qk−1 + qk−2 shows that the numbers qk increase strictly with k, so ck − ck−1 → 0 as k → ∞. Hence y = z, and the whole sequence converges. We define the limit of the sequence of convergents to be the value of the infinite continued fraction [a0 ; a1 , a2 , . ]. For example, if xn = [2; 2, 2, . . , 2] (with n + 1 2s), then we have xn = 2 + 1 xn−1 . If limn→∞ xn = u, then clearly 1 u = 2+ , u √ 2 so u√− 2u − 1 = 0, or u = 1 ± 2. But u is obviously positive; so we have u = 1 + 2.

We have p0 = a0 , p1 = a0 a1 + 1, q0 = 1, q1 = a1 , and so p1 q0 −q1 p0 = 1 = (−1)0 , so the induction starts. If we assume that pk−1 qk−2 − qk−1 pk−2 = (−1)k−2 , then we have pk = ak pk−1 + pk−2 and qk = ak qk−1 = qk−2 ; so pk qk−1 − qk pk−1 = = = = (ak pk−1 + pk−2 )qk−1 − (ak qk−1 + qk−2 )pk−1 pk−2 qk−1 − qk−2 pk−1 −(−1)k−2 (−1)k−1 . 26 CHAPTER 3. FINITE CONTINUED FRACTIONS (b) Divide both sides of (a) by qk−1 qk and use the fact that ck = pk /qk and ck−1 = pk−1 /qk−1 . 8 The convergents satisfy c0 < c2 < c4 < · · · < c5 < c3 < c1 .

An−1 , an ], then −1/y = [an ; an−1 , . . , a1 , a0 ]. Proof y = [a0 ; a1 , . . , an , y] = ypn + pn−1 , yqn + qn−1 so qn y2 + (qn−1 − pn )y + pn−1 = 0. Let z = [an ; an−1 , . . , a0 ]. Then z[an , . . , a0 ] + [an , . . , a1 ] z[an−1 , . . , a0 ] + [an−1 , . . , a1 ] zpn + qn = , zpn−1 + qn−1 z = [an ; an−1 , . . , a0 , z] = where we use the fact that [a0 , . . , an ] = [an , . . , a0 ]. So pn−1 z2 + (qn−1 − pn )z − qn = 0. In other words, qn (−1/z)2 + (qn−1 − pn )(−1/z) + pn−1 = 0.