480 Real Analysis
withx 1 =
√
7 andx 2 =√
7 −
√
- Let us first determine the possible values of the limit
L, assuming that it exists. Passing to the limit in the recurrence relation, we obtain
L=
√
7 −
√
7 +L.
Squaring twice, we obtain the polynomial equationL^4 − 14 L^2 −L+ 42 =0. Two roots
are easy to find by investigating the divisors of 42, and they areL=2 andL=−3. The
other two areL=^12 ±
√ 29
2. Only the positive roots qualify, and of them1
2 +√ 29
2 is not
a root of the original equation, since
1
2
+
√
29
2
> 3 >
√
7 −
√
7 + 3 >
√√
√√
7 −
√
7 +
1
2
+
√
29
2
.
So the only possible value of the limit isL=2.
Letxn= 2 +αn. Thenα 1 ,α 2 ∈( 0 , 1 ). Also,
αn+ 2 =3 −
√
9 +αn
√
7 −√
9 +αn+ 4.
Ifαn∈( 0 , 1 ), then
0 >αn+ 2 >3 −
√
9 +αn
4≥−
1
2
αn,where the last inequality follows from 3+ 2 αn≥
√
9 +αn. Similarly, ifαn∈(− 1 , 0 ),
then
0 <αn+ 2 <3 −
√
9 +αn
4≤
1
2
|αn|,where the last inequality follows from 3<
√
9 −|αn|+ 2 α. Inductively, we obtain
thatαn∈(− 2 −n/^2 , 2 −n/^2 ), and henceαn→0. Consequently, the sequence(xn)nis
convergent, and its limit is 2.
(13th W.L. Putnam Mathematics Competition, 1953)
340.The solution is a direct application of the Cesàro–Stolz theorem. Indeed, if we let
an=lnunandbn=n, then
lnun+ 1
un=lnun+ 1 −lnun=an+ 1 −an
bn+ 1 −bnand
lnn√
un=1
nlnun=an
bn