数列不等式的证明技巧

数列不等式证明技巧 遂宁市第七中学

例1.已知an?1?1?郑廷楷

211,a1?,求证(-1)a1 +(-1)2a2+…+(-1)nan<1 an7证明:先证0?a2n?1?2,a2n?2(*)

1125<2,a2??2 711∴当n=1时(*)式成立

当n=1时∵0

a2n?12(a2n?2)(a2n?1)

a2n(a2n?2)3a2n?1??423?4a2n?12554注意a2?,b1?a2?a1?

1177bbb54bn?b1?2?3??n??(2?3)n?1

b1b2bn?177bn?12(a2n?1)?=bn1a2n(a2n?2)2?2?2-3

77[1?(2?3)]因此当n为偶数时:(-1)a1 +(-1)2a2+…+(-1)nan<1成立 当n为偶奇数时多了个负加数因此原式也成立 故原命题总成立。 例2.数列{an}中,an?1?1?a,求证:对于一切正整数n都有an?1 anb1?b2?b3???bn?54?4(3?1)<1 11

1?a211?a1?a3证明:a1?1?a?,a2??a? ?a?221?aa11?a1?a1?an?111?a21?a4,猜出:an?(*) a3??a??a?1?ana21?a31?a3下面用数数归纳法证明(*)

当n=1时(*)显然成立

1?ak?1假设当n=k时(*)成立,即ak?

1?ak11?ak1?ak?2则ak?1??a? ?a?ak1?ak?11?ak?1即当n=k+1时(*)也成立

1?an?1故当n?N?时an? 成立

1?an1?an?11?an?1?1?anan(1?a)an?1??1???0,?an?1 nnn1?a1?a1?a

例3.证明: a?1,n?2,且n?N?,求证an?11?n(a?) naa1n1a(a2n?1)a(1?a2n)a?n2证明:(a?a3???a2n?1) ?n?n21a(a?1)a(1?a)aa?aan?211?nn(a?a3???a2n?1)n=nn(an1)n?n aa11111?0,于是an?n?n(a?) aaa1例4.数列bn?3n?1,而数列an=loga(1?) (a大于1),Sn是数列an前n项之和,

bnlogabn?1求证:Sn?

3113n)?loga证明:an?loga(1?)?loga(1?

bn3n?13n?1363nSn?a1?a2???an=loga?loga???loga

253n?1363n) =loga(????253n?1由假分数的性质得

因a?1,故a?

363n473n?1?????(1) ????253n?1363n363n583n?2?????(2) ????253n?1473n?1363n363n又????=????(3) 253n?1253n?1(1)×(2)×(3)得 363n33?4?5???(3n?2)(3n?1)(3n?2)?(???? )?2?3?4???(3n?1)2?3253n?1363n3(3n?1)(3n?2)当n≥2 时(????>3n+2=bn?1 )?253n?12?3因为a?1

logabn?1363n故有3loga(???? )?logabn?1,即Sn?3253n?1

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