Real Analysis 479xn+ 1 −( 1 +√
2 )=
√
1 + 2 xn−√
1 + 2 ( 1 +
√
2 )=
2 (xn−( 1 −√
2 ))
√
1 + 2 xn+√
1 + 2 ( 1 +
√
2 )
<
xn−( 1 +√
2 )
1 +
√
2
.
From here we deduce that
x 1969 −( 1 +√
2 )<
50
( 1 +
√
2 )^1968
< 10 −^3 ,
and the approximation ofx 1969 with two decimal places coincides with that of 1+
√
2 =
2 .41. This argument proves also that the limit of the sequence is 1+
√
2.
(St. Petersburg Mathematical Olympiad, 1969)338.Write the equation as
√
x+ 2√
x+···+ 2√
x+ 2√
x+ 2 x=x.We can iterate this equality infinitely many times, always replacing the very lastxby its
value given by the left-hand side. We conclude thatxshould satisfy
√
x+ 2
√
x+ 2√
x+ 2 ··· =x,provided that the expression on the left makes sense! Let us check that indeed the
recursive sequence given byx 0 =x, andxn+ 1 =
√
x+ 2 xn,n≥0, converges for any
solutionxto the original equation. Squaring the equation, we find thatx<x^2 , hence
x>1. But thenxn+ 1 <xn, because it reduces tox^2 n− 2 xn+x>0. This is always
true, since when viewed as a quadratic function inxn, the left-hand side has negative
discriminant. Our claim is proved, and we can now transform the equation, the one with
infinitely many square roots, into the much simpler
x=√
x+ 2 x.This has the unique solutionx =3, which is also the unique solution to the equation
from the statement, and this regardless of the number of radicals.
(D.O. Shklyarski, N.N. Chentsov, I.M. Yaglom,Selected Problems and Theorems in
Elementary Mathematics, Arithmetic and Algebra, Mir, Moscow)
339.The sequence satisfies the recurrence relation
xn+ 2 =√
7 −
√
7 +xn,n≥ 1 ,