3.1 Sequences and Series 115L−
2
+
(
−L
yn 0
ym+
2
·
yn 0
ym+
xn 0
ym)
<
xm
ym<L+
2
+
(
−L
yn 0
ym−
2
·
yn 0
ym+
xn 0
ym)
.
Becauseyn→∞, there existsn 1 >n 0 such that form≥n 1 , the absolute values of the
terms in the parentheses are less than 2. Hence form≥n 1 ,
L−<
xm
ym<L+.
Sincewas arbitrary, this proves that the sequence(xynn)nconverges toL.
We continue this discussion with an application to Cesàro means. By definition, the
Cesàro means of a sequence(an)n≥ 1 are
sn=
a 1 +a 2 +···+an
n,n≥ 1.Theorem.If(an)n≥ 1 converges toL, then(sn)n≥ 1 also converges toL.
Proof.Apply the Cesàro–Stolz theorem to the sequencesxn=a 1 +a 2 +···+anand
yn=n,n≥1.
The Cesàro–Stolz theorem can be used to solve the following problems.340.If(un)nis a sequence of positive real numbers and if limn→∞uun+n^1 =u>0, then
limn→∞n
√
un=u.341.Letpbe a real number,p =1. Compute
nlim→∞1 p+ 2 p+···+np
np+^1.
342.Let 0<x 0 <1 andxn+ 1 =xn−xn^2 forn≥0. Compute limn→∞nxn.
343.Letx 0 ∈[− 1 , 1 ]and xn+ 1 = xn−arcsin(sin^2 xn) forn ≥ 0. Compute
limn→∞
√
nxn.344.For an arbitrary numberx 0 ∈ ( 0 ,π)define recursively the sequence(xn)nby
xn+ 1 =sinxn,n≥0. Compute limn→∞
√
nxn.345.Letf:R→Rbe a continuous function such that the sequence(an)n≥ 0 defined
byan =
∫ 1
0 f(n+x)dxis convergent. Prove that the sequence(bn)n≥^0 , with
bn=∫ 1
0 f (nx)dxis also convergent.346.Consider the polynomialP(x)=amxm+am− 1 xm−^1 + ··· +a 0 ,ai >0,i =
0 , 1 ,...,m. Denote byAnandGnthe arithmetic and, respectively, geometric
means of the numbersP( 1 ), P ( 2 ),...,P(n). Prove that
nlim→∞An
Gn=
em
m+ 1