各类平均与数列极限

已知数列${a_n},{b_n}$满足$a_0=a,b_0=b(a>0,b>0,a eq b)$,且
[egin{cases}
a_{n+1}=frac{1}{2}(a_n+b_n)\
b_{n+1}=sqrt{a_{n+1}b_n}.
end{cases}(n=0,1,2,ldots)]
试求${a_n},{b_n}$的通项公式以及它们的极限.

由题设条件有
[left( frac{a_{n+1}}{b_{n+1}} ight) ^2=frac{a_{n+1}^{2}}{a_{n+1}b_n}=frac{a_{n+1}}{b_n}=frac{1}{2}left( frac{a_n}{b_n}+1 ight).]
令$t_n=frac{a_n}{b_n} (n=0,1,2,ldots)$,则
[t_{n+1}^2=frac{1}{2}(t_n+1),quad t_0=frac{a}{b}.]
再令$t_n=frac{1}{2}x_n$,则
[x_{n+1}^2=x_n+2,quad x_0=2frac{a}{b}.]
由方程$x_0=t+frac{1}{t}=2frac{a}{b}$,解得$t_{1,2}=frac{apmsqrt{a^2-b^2}}{b}$.

易知,当$x_0>2$时, $x_n>2$;当$0<x_0<2$时, $0<x_n<2$.因此
[
x_n=egin{cases}
2cos left( frac{1}{2^n}arccos frac{a}{b} ight) ,& 0<a<b;\
left( frac{a+sqrt{a^2-b^2}}{b} ight) ^{1/2^n}+left( frac{a-sqrt{a^2-b^2}}{b} ight) ^{1/2^n},& 0<b<a.\
end{cases}
]
又因为$frac{b_{n+1}}{b_n}=frac{a_{n+1}}{b_{n+1}}=frac{1}{2}x_{n+1}$,所以,当$0<a<b$时,
egin{align*}
b_n&=b_0frac{x_1x_2cdots x_n}{2^n}=bcos frac{ heta}{2}cos frac{ heta}{2^2}cdots cos frac{ heta}{2^n}
\
&=frac{bsin heta}{2^nsin frac{ heta}{2^n}},quad heta =arccos frac{a}{b},n=0,1,2,ldots
end{align*}
则
[a_n=frac{bsin heta}{2^n}cot frac{ heta}{2^n}]
并且[lim_{n ightarrow infty}a_n=lim_{n ightarrow infty}b_n=frac{bsin heta}{ heta}=frac{sqrt{b^2-a^2}}{arccos frac{a}{b}}.]

当$a>b>0$时,
egin{align*}
x_n&=left( frac{a+sqrt{a^2-b^2}}{b} ight) ^{1/2^n}+left( frac{a-sqrt{a^2-b^2}}{b} ight) ^{1/2^n}
\
&=e^{alpha /2^n}+e^{-alpha /2^n}=2mathrm{cosh}left( frac{alpha}{2^n} ight),quad alpha=lnfrac{a+sqrt{a^2-b^2}}{b}
end{align*}
其中$mathrm{cosh} x=frac{1}{2}left(e^x+e^{-x} ight)$为双曲余弦函数,注意到$mathrm{sinh}x=2mathrm{sinh}frac{x}{2}mathrm{cosh}frac{x}{2}$,其中$mathrm{sinh} x=frac{1}{2}left(e^x-e^{-x} ight)$为双曲正弦函数,我们有
[b_n=b_0frac{x_1x_2cdots x_n}{2^n}=bmathrm{cosh}frac{alpha}{2}mathrm{cosh}frac{alpha}{2^2}cdots mathrm{cosh}frac{alpha}{2^n}=frac{bmathrm{sinh}alpha}{2^nmathrm{sinh}frac{alpha}{2^n}},]
故
[a_n=frac{bmathrm{sinh}frac{alpha}{2}}{2^n}mathrm{coth} frac{alpha}{2^n},]
并且
[
lim_{n ightarrow infty}a_n=lim_{n ightarrow infty}b_n=frac{bmathrm{sinh} alpha}{alpha}=frac{sqrt{a^2-b^2}}{ln frac{a+sqrt{a^2-b^2}}{b}}.
]

(1990年匈牙利奥赛)设$a_0=1,a_n=frac{sqrt{1+a_{n-1}^2}-1}{a_{n-1}}$,则$a_n>frac{pi}{2^{n+2}}$.

设$a_kgeq 0,a_{n+m}leq a_n+a_m$,则对任意$ngeq m$,均有
[a_nle ma_1+left( frac{n}{m}-1 ight) a_m.]
特别地,若$a_1=1,a_n>1(ngeq 2)$且$a_{n+m}leq a_n+a_m$,则$a_n<n$.

(1989年30届IMO预选题)定义数列${a_n},{b_n}$如下: $a_0=frac{sqrt{2}} {2},b_0=1$,
[ a_{n+1}=a_0sqrt{1-sqrt{1-a_n^2}},quad b_{n+1}=frac{sqrt{1+b_n^2}-1}{b_n}(n=0,1,2,ldots)]
求证:对每一个$n=0,1,2,ldots$,有不等式
[2^{n+2}a_n<pi<2^{n+2}b_n.]

提示:令$b_{n}= an heta_n$,可求得$ heta_n=frac{pi}{2^{n+2}}$,故$b_n= anfrac{pi} {2^{n+2}}$.再利用不等式$sin x<x< an x(0<x<pi/2)$即可.

(叶军P 283)已知数列${a_n},{b_n}$满足$a_0=a,b_0=b(a>0,b>0)$,且
[egin{cases}
a_{n+1}=frac{1}{2}(a_n+b_n)\
b_{n+1}=frac{2a_nb_n}{a_n+b_n}.
end{cases}(n=0,1,2,ldots)]
试求${a_n},{b_n}$的通项公式.

易证$a_nb_n=a_0b_0=ab(ngeq 0)$以及
[a_{n+1}=frac{1}{2}left( a_n+frac{ab}{a_n} ight).]
因此
[
frac{a_{n+1}-sqrt{ab}}{a_{n+1}+sqrt{ab}}=left( frac{a_n-sqrt{ab}}{a_n+sqrt{ab}} ight) ^2.
]
解得
egin{align*}
a_n&=frac{left( a+sqrt{ab} ight) ^{2^n}+left( a+sqrt{ab} ight) ^{2^n}}{left( a+sqrt{ab} ight) ^{2^n}-left( a-sqrt{ab} ight) ^{2^n}}sqrt{ab},
\
b_n&=frac{left( a+sqrt{ab} ight) ^{2^n}-left( a+sqrt{ab} ight) ^{2^n}}{left( a+sqrt{ab} ight) ^{2^n}+left( a-sqrt{ab} ight) ^{2^n}}sqrt{ab}.
end{align*}

已知数列${a_n},{b_n}$满足$a_0=a,b_0=b(a>0,b>0)$,且
[egin{cases}
a_{n+1}=frac{2a_nb_n}{a_n+b_n}\
b_{n+1}=sqrt{a_{n+1}b_n}.
end{cases}(n=0,1,2,ldots)]
试求${a_n},{b_n}$的通项公式及极限.

由$b_{n+1}=sqrt{a_{n+1}b_n}$可知
[a_{n+1}=frac{b_{n+1}^{2}}{b_n}.]
由$a_{n+1}=frac{2a_nb_n}{a_n+b_n}$可知
[frac{b_{n+1}^{2}}{b_n}=frac{2frac{b_{n}^{2}}{b_{n-1}}b_n}{frac{b_{n}^{2}}{b_{n-1}}+b_n}=frac{2b_{n}^{2}}{b_n+b_{n-1}},]
故
[
left( frac{b_n}{b_{n+1}} ight) ^2=frac{1}{2}left( 1+frac{b_{n-1}}{b_n} ight).
]
令$x_n=frac{b_{n-1}}{b_n}$,则$x_{n+1}^2=frac{1}{2}left( 1+x_n ight)$且$x_1=frac{b_0}{b_1}=sqrt{frac{a+b}{2a}}$.

若$0<bleq a$,则$0<x_1leq 1$,进一步可知$0<x_nleq 1$.令$x_n=cos2 heta_n$,则$x_{n+1}=sqrt{frac{1}{2}left( 1+x_n ight)}=cos heta_n$,则$ heta_{n+1}=frac{1}{2} heta_n$,则
[ heta _n=frac{1}{2^n}arccos sqrt{frac{a+b}{2a}},quad x_n=cos left( frac{1}{2^{n-1}}arccos sqrt{frac{a+b}{2a}} ight).]
因此
egin{align*}
b_n&=b_0Bigg/prod_{k=1}^n{cos left( frac{1}{2^{k-1}}arccos sqrt{frac{a+b}{2a}} ight)}=bBigg/frac{sin left( ext{2}arccos sqrt{frac{a+b}{2a}} ight)}{2^nsin left( frac{1}{2^{n-1}}arccos sqrt{frac{a+b}{2a}} ight)}
\
&=frac{2^nsin left( frac{1}{2^{n-1}}arccos sqrt{frac{a+b}{2a}} ight)}{sin left( ext{2}arccos sqrt{frac{a+b}{2a}} ight)}b,
end{align*}
而
[
a_n=frac{b_{n}^{2}}{b_{n-1}}=frac{2^n an left( frac{1}{2^{n-1}}arccos sqrt{frac{a+b}{2a}} ight)}{sin left( ext{2}arccos sqrt{frac{a+b}{2a}} ight)}b.
]
特别地,若$a=2sqrt{3},b=3$,利用$arccosfrac{sqrt{6}+sqrt{2}}{4}=frac{pi}{12}$可知
[
a_n=3cdot 2^{n+1} an frac{pi}{3cdot 2^{n+1}},qquad b_n=3cdot 2^{n+1}sin frac{pi}{3cdot 2^{n+1}}.
]
此时$lim_{n ightarrow infty}a_n=lim_{n ightarrow infty}b_n=pi$.

若$bgeq a>0$,则令$x_n=frac{1}{2}left( t_n+frac{1}{t_n} ight)$,则
[x_{n+1}=sqrt{frac{1}{2}left( 1+x_n ight)}=frac{1}{2}left( sqrt{t_n}+frac{1}{sqrt{t_n}} ight),]
因此$t_{n+1}=sqrt{t_n}$,由$x_1=frac{1}{2}left( t_1+frac{1}{t_1} ight) =sqrt{frac{a+b}{2a}}$可知
[t_1=left( sqrt{frac{a+b}{2a}}pmsqrt{frac{b-a}{2a}} ight) ^{1/2^{n-1}}.]
因此
[
x_n=frac{1}{2}left[ left( sqrt{frac{a+b}{2a}}+sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}}+left( sqrt{frac{a+b}{2a}}-sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}} ight],
]
则
egin{align*}
b_n&=frac{2^{n-1}ab}{sqrt{b^2-a^2}}left[ left( sqrt{frac{a+b}{2a}}+sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}}-left( sqrt{frac{a+b}{2a}}-sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}} ight]
\
a_n&=frac{2^nab}{sqrt{b^2-a^2}}frac{left( sqrt{frac{a+b}{2a}}+sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}}-left( sqrt{frac{a+b}{2a}}-sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}}}{left( sqrt{frac{a+b}{2a}}+sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}}+left( sqrt{frac{a+b}{2a}}-sqrt{frac{b-a}{2a}} ight) ^{ ext{1/}2^{n-1}}}.
end{align*}

---

(叶军P 310)设${a_n}$为有下列性质的数列:
[1=a_0leq a_1leq a_2leqcdotsleq a_nleq cdots ag{1}]
又${b_n}$是由下式定义的数列
[b_n=sum_{k=1}^n{left( 1-frac{a_{k-1}}{a_k} ight) cdot frac{1}{sqrt{a_k}}},quad n=1,2,ldots ag{2}]
证明: (a)对所有$n=1,2,3,ldots$,有$0leq b_n<2$;

(b)对$0leq c<2$的任一$c$,总存在一个具有性质(1)的数列${a_n}$,使得由(2)导出的数列${b_n}$中有无限多个下标$n$满足$b_n>c$.

因为
egin{align*}
left( 1-frac{a_{k-1}}{a_k} ight) cdot frac{1}{sqrt{a_k}}&=frac{a_{k-1}}{sqrt{a_k}}left( frac{1}{a_{k-1}}-frac{1}{a_k} ight)
\
&=frac{a_{k-1}}{sqrt{a_k}}left( frac{1}{sqrt{a_{k-1}}}+frac{1}{sqrt{a_k}} ight) left( frac{1}{sqrt{a_{k-1}}}-frac{1}{sqrt{a_k}} ight)
\
&=left( sqrt{frac{a_{k-1}}{a_k}}+frac{a_{k-1}}{a_k} ight) left( frac{1}{sqrt{a_{k-1}}}-frac{1}{sqrt{a_k}} ight) <2left( frac{1}{sqrt{a_{k-1}}}-frac{1}{sqrt{a_k}} ight),
end{align*}
所以
egin{align*}
0&le b_n=sum_{k=1}^n{left( 1-frac{a_{k-1}}{a_k} ight) cdot frac{1}{sqrt{a_k}}}
\
&<sum_{k=1}^n{2left( frac{1}{sqrt{a_{k-1}}}-frac{1}{sqrt{a_k}} ight)}<frac{2}{sqrt{a_0}}=2.
end{align*}

令$frac{1}{sqrt{a_k}}=d^k$,因为当$0<d<1$时,条件(1)满足,又和式$b_n$中第$k$项是
[left( 1-frac{d^{-2left( k-1 ight)}}{d^{-2k}} ight) d^k=left( 1-d^2 ight) d^k,]
因此
egin{align*}
b_n&=sum_{k=1}^n{left( 1-d^2 ight) d^k}=left( 1-d^2 ight) sum_{k=1}^n{d^k}
\
&=left( 1-d^2 ight) frac{d-d^{n+1}}{1-d}=dleft( 1+d ight) left( 1-d^n ight).
end{align*}
现在要求对无穷多个$n$, 均有$dleft( 1+d ight) left( 1-d^n ight)>c$,即
[d^n<1-frac{c}{d(1+d)}. ag{3}]
对充分大的$n$, $d^n o 0$,故只需
[0<1-frac{c}{d(1+d)},quad d(1+d)>c.]
故只需$d(1+d)>2d^2>c$.因此,只需选择$d$,满足$sqrt{frac{c}{2}}<d<1$.这时有$d(1+d)>c$,故(3)式右端为正数,因为$din (0,1),d^n o 0$,所以存在充分大的自然数$M$,使得当$n>M$时(3)式成立,于是(b)得证.