482 Real Analysis
then this will equal the square of the limit under discussion. We use the Cesàro–Stolz
theorem.
Suppose 0<x 0 ≤1 (the casesx 0 <0 andx 0 =0 being trivial; see above). If
0 <xn ≤1, then 0<arcsin(sin^2 xn)<arcsin(sinxn)=xn,so0<xn+ 1 <xn.It
follows by induction onnthatxn∈( 0 , 1 ]for allnandxndecreases to 0. Rewriting the
recurrence as sinxn+ 1 =sinxn
√
1 −sin^4 xn−sin^2 xncosxngives1
sinxn+ 1−
1
sinxn=
sinxn−sinxn+ 1
sinxnsinxn+ 1=sinxn−sinxn√
1 −sin^4 xn+sin^2 xncosxn
sinxn(sinxn√
1 −sin^4 xn−sin^2 xncosxn)=1 −
√
1 −sin^4 xn+sinxncosxn
sinxn√
1 −sin^4 xn−sin^2 xncosxn=
sin^4 xn
1 +√
1 −sin^4 xn
+sinxncosxnsinxn√
1 −sin^4 xn−sin^2 xncosxn=
sin^3 xn
1 +√
1 −sin^4 xn+cosxn
√
1 −sin^4 xn−sinxncosxn.
Hence
lim
n→∞(
1
sinxn+ 1−
1
sinxn)
= 1.
From the Cesàro–Stolz theorem it follows that limn→∞nsin^1 xn =1, and so we have
limn→∞nxn=1.
(Gazeta Matematic ̆a (Mathematics Gazette, Bucharest), 2002, proposed by
T. Andreescu)
344.We compute the square of the reciprocal of the limit, namely limn→∞nx^12 n. To this
end, we apply the Cesàro–Stolz theorem to the sequencesan=x^12 nandbn=n. First,
note that limn→∞xn=0. Indeed, in view of the inequality 0<sinx<xon( 0 ,π), the
sequence is bounded and decreasing, and the limitLsatisfiesL=sinL,soL=0. We
then have
lim
n→∞(
1
x^2 n+ 1−
1
xn^2)
=lim
n→∞(
1
sin^2 xn−
1
xn^2)
=lim
n→∞xn^2 −sin^2 xn
x^2 nsin^2 xn=lim
xn→ 0xn^2 −^12 ( 1 −cos 2xn)
1
2 x
n^2 (^1 −cos 2xn) =xlimn→^02 xn^2 −[
( 2 xn)^2
2! −( 2 xn)^4
4! +···]
xn^2[
( 2 xn)^2
2! −( 2 xn)^4
4! +···