476 Real Analysis
This proves that the sequences are Cauchy, hence convergent. Because asntends to
infinity (an,bn,cn)approaches 0, the three sequences converge to the same limitL.
Finally, because for alln,an+bn+cn=a 0 +b 0 +c 0 , we should have 3L=a 0 +b 0 +c 0 ;
hence the common limit is(a^0 +b 30 +c^0 ).
329.Because
∑
anconverges, Cauchy’s criterion implies thatlim
n→∞
(an/ 2 + 1 +an/ 2 + 2 +···+an)= 0.By monotonicity
an/ 2 + 1 +an/ 2 + 2 +···+an≥⌈n
2⌉
an,so limn→∞n 2 an=0. Consequently, limn→∞n 2 an=0, and hence limn→∞nan=0, as
desired.
(Abel’s lemma)
330.Think of the larger map as a domainDin the plane. The change of scale from one
map to the other is a contraction, and since the smaller map is placed inside the larger,
the contraction mapsDtoD. Translating into mathematical language, a point such as
the one described in the statement is a fixed point for this contraction. And by the fixed
point theorem the point exists and is unique.
331.Define the functionf(x)=sinx+t. Then for any real numbersx 1 andx 2 ,
|f(x 1 )−f(x 2 )|=||·|sinx 1 −sinx 2 |≤ 2 ||·∣∣
∣∣sinx^1 −x^2
2∣∣
∣∣·
∣∣
∣∣cosx^1 +x^2
2∣∣
∣∣
≤ 2 ||·
∣∣
∣∣sinx^1 −x^2
2∣∣
∣∣≤|x 1 −x 2 |.Hencefis a contraction, and there exists a uniquexsuch thatf(x)=sinx+t=x.
Thisxis the unique solution to the equation.
(J. Kepler)
332.Definef :( 0 ,∞) →( 0 ,∞),f(x)=^12 (x+xc). Thenf′(x)=^12 ( 1 −xc 2 ),
which is negative forx<
√
cand positive forx>√
c. This shows that√
cis a global
minimum forfand henceforthf(( 0 ,∞))⊂[
√
c,∞). Shifting indices, we can assume
thatx 0 ≥
√
c. Note that|f′(x)|<^12 forx∈[√
c,∞),sofis a contraction on this
interval. Becausexn =f(f(···f(x 0 )),n≥1, the sequence(xn)nconverges to the
unique fixed pointx∗off. Passing to the limit in the recurrence relation, we obtain
x∗=^12 (x∗+xc∗), which is equivalent to the quadratic equation(x∗)^2 −c=0. We obtain
the desired limit of the sequencex∗=
√
c.
(Hero)