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n+1 1+An

(1) a(n+1)=an/(an+1) 1/a(n+1) = (an+1)/an 1/a(n+1) -1/an = 1 =>(1/an)是等差数列 1/an -1/a1= n-1 1/an =n an =1/n (2) bn =1/(2^n.an) = (1/2)[n(1/2)^(n-1)] consider 1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1) 1+2x+..+nx^(n-1) =[(x^(n+1)- ...

a(n+1)=an+1/n-1/(n+1) a(n+1)+1/(n+1)=an+1/n 设bn=an+1/n 则:b(n+1)=bn b1=a1+1/1=2 所以:数列bn为常数列bn=2 所以an+1/n=2 an=2-1/n

a[n+1]-a[n] > 1/4 1/(1-a[n]) - a[n] = (1-2a[n])^2/(4-4a[n]) > 0 a[n]单增 又a[n]=1/4 (极限时通常严格不等式要变成非严格不等式) (A-1/2)^2

【方法一】显然an≥1,从而an+1≥2,(n=1,2,3,…).因为|an+1?an|=|1+an?1+an?1|=11+an+1+an?1|an?an?1|≤12|an?an?1|,(n=2,3,…),所以{an}是压缩数列,从而{an}收敛,设limn→∞an=a,则a≥2.因为an+1=1+an,令n→∞可得,a=1+a,从而a2-...

∵a(n+1)-an=an/(n+1) ∴a(n+1)=an+an/(n+1) =an*(n+2)/(n+1) ∴a(n+1)/an=(n+2)/(n+1) 那么an/a(n-1)=(n+1)/n a(n-1)/a(n-2)=n/(n-1) ………………………… a3/a2=4/3 a2/a1=3/2 累乘,得:an/a1=(n+1)/2 而a1=1,∴an=(n+1)/2

a(n+1)/an=n/n+1 则an/a(n-1)=(n-1)/n a(n-1)/a(n-2)=(n-2)/(n-1) .......... a2/a1=1/2 叠乘 an/a1=1/n 因a1=1 故an=1/n

因为a(n+1)=an-n,所以a(n+1)-an=-n 所以 an - a(n-1) = -(n-1) a(n-1 )-a(n-2)= -(n-2) ...... ...... a2 - a1 = -1 等式相加得 an-a1= - (n-1+n-2+...+1) an= - (n-1+n-2+...+1)+a1= - (n-1) /2 + 1 =(n^2-n)/2 + 1

an/a(n-1)=[n-1]/[n+1] a2/a1=1/3 a3/a2=2/4 a4/a3=3/5 . . . a(n-1)/a(n-2)=n-2/n an/a(n-1)=n-1/n+1 相乘:an/a1=2/n*(n+1) an=2/n*(n+1)=2*[1/n-1/(n+1)] Sn=2[1-1/2+1/2-1/3+...+1/n-1/(n+1)]=2[1-1/(n+1)]=2n/(n+1)

易证an>0,即0是一个下界 a1=2>1① 假设ak>1,那么ak+1=1/2*(ak+1/ak)≥1/2*2√(ak*1/ak)=1,当且仅当ak=1/ak时取等号,即ak=1时,这和ak>1矛盾 ∴上述不等式取不到等号,即从ak>1可推出ak+1>1② 综合①②得an>1 an+1-an=1/2*(an+1/an)-an=1/2*(-an+1/an)=(1-...

an=1+2+3+……+n =n(1+n)/2 1/an=2/[n(n+1)] =2[1/n - 1/(n+1)] 数列{1/an}的前n项和Sn=a1+a2+a3+……+an =2[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)] =2[1-1/(n+1)] =2n/(n+1) Sn=2n/(n+1) (n为正整数)

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