Advanced book on Mathematics Olympiad

Real Analysis 477

333.Define

xn=

√

1 +

√

1 +

√

1 +···+

√

1 ,n≥ 1 ,

where in this expression there arensquare roots. Note thatxn+ 1 is obtained fromxnby
replacing

√

1by

√

1 +

√

1 at the far end. The square root function being increasing, the
sequence(xn)nis increasing. To prove that the sequence is bounded, we use the recurrence
relationxn+ 1 =

√

1 +xn,n≥1. Then fromxn<2, we obtain thatxn+ 1 =

√

√^1 +xn<
1 + 2 <2, so inductivelyxn<2 for alln. Being bounded and monotonic, the sequence
(xn)nis convergent. LetLbe its limit (which must be greater than 1). Passing to the
limit in the recurrence relation, we obtainL=

√

1 +L,orL^2 −L− 1 =0. The only
positive solution is the golden ratio

√ 5 + 1

2 , which is therefore the limit of the sequence.

334.If the sequence converges to a certain limitL, thenL=

√

a+bL,soLis equal to
the (unique) positive rootαof the equationx^2 −bx−a=0.
The convergence is proved by verifying that the sequence is monotonic and bounded.
The conditionxn+ 1 ≥xntranslates tox^2 n≥a+bxn, which holds if and only ifxn>α.
On the other hand, ifxn≥α, thenxn^2 + 1 =a+bxn≥a+bα=α^2 ; hencexn+ 1 ≥α.
Similarly, ifxn≤α, thenxn+ 1 ≤α. There are two situations. Eitherx 1 <α, and then
by inductionxn<αfor alln, and hencexn+ 1 >xnfor alln. In this case the sequence
is increasing and bounded from above byα; therefore, it is convergent, its limit being of
courseα.Orx 1 ≥α, in which case the sequence is decreasing and bounded from below
by the sameα, and the limit is againα.

335.By the AM–GM inequality,an<bn,n≥1. Also,

an+ 1 −an=

√

anbn−an=

√

an(

√

bn−

√

an)> 0 ;

hence the sequence(an)nis increasing. Similarly,

bn+ 1 −bn=

an+bn

2

−bn=

an−bn

2

< 0 ,

so the sequencebnis decreasing. Moreover,

a 0 <a 1 <a 2 <···<an<bn<···<b 1 <b 0 ,

for alln, which shows that both sequences are bounded. By the Weierstrass theorem,
they are convergent. Leta=limn→∞anandb=limn→∞bn. Passing to the limit in the
first recurrence relation, we obtaina=

√

ab, whencea=b. Done.

Remark.The common limit, denoted byM(a,b), is called the arithmetic–geometric
mean of the numbersaandb. It was Gauss who first discovered, as a result of laborious