Advanced book on Mathematics Olympiad

3.1 Sequences and Series 113

1

√

2

(x+ 1 )≤f(x)≤

√

2 (x+ 1 ).

Repeating successively the argument, we find that

2 −

21 n

(x+ 1 )≤f(x)≤ 2

21 n

(x+ 1 ), forn≥ 1.

If in this double inequality we letn→∞, we obtainx+ 1 ≤f(x)≤x+1, and hence
f(x)=x+1. The particular casex=2 yields Ramanujan’s formula
√

1 + 2

√

1 + 3

√

1 + 4

√

1 +··· = 3. 

Here are some problems of this kind.

333.Compute
√

1 +

√

1 +

√

1 +

√

1 +···.

334.Letaandbbe real numbers. Prove that the recurrence sequence(xn)ndefined by
x 1 >0 andxn+ 1 =

√

a+bxn,n≥1, is convergent, and find its limit.

335.Let 0<a<bbe two real numbers. Define the sequences(an)nand(bn)nby
a 0 =a,b 0 =b, and

an+ 1 =

√

anbn,bn+ 1 =

an+bn

2

,n≥ 0.

Prove that the two sequences are convergent and have the same limit.

336.Prove that forn≥2, the equationxn+x− 1 =0 has a unique root in the interval
[ 0 , 1 ].Ifxndenotes this root, prove that the sequence(xn)nis convergent and find
its limit.

337.Compute up to two decimal places the number
√√
√
√
1 + 2

√

1 + 2

√

1 +···+ 2

√

1 + 2

√

1969 ,

where the expression contains 1969 square roots.

338.Find the positive real solutions to the equation
√

x+ 2

√

x+···+ 2

√

x+ 2

√

3 x=x.