Question

数列an=4^n (-2)^n，求证1/a1+1/a2+1/a3+……+1/an<7/12

an=4^n +(-2)^n

Answer 1

算出数列的前几项得，a1=2，a2=20,a3=56,a4=272，4^n+(-2)^n>=4^n-2^n=2^n(2^n-1),所以1/an <=1/2^n(2^n-1)=1/(2^n-1)-1/2^n。所以1/a1+1/a2+...1/an<1/2+1/20+1/56+1/272+(1/(2^5-1)-1/2^5+......1/(2^n-1)-1/2^n)。下面估算(1/(2^5-1)-1/2^5+......1/(2^n-1)-1/2^n)的值，
令(1/(2^5-1)-1/2^5+......1/(2^n-1)-1/2^n)=t，则(1/(2^5-1)-1/2^5+......1/(2^n-1)-1/2^n))<(1/(2^5-2)-1/2^5+.....1/(2^n-2)-1/2^n)-1/2^n)=1/2(1/(2^4-1)-1/2^4+......1/(2^(n-1)-1)-1/2^(n-1))=1/2(1/(2^4-1)-1/2^4)+1/2(1/(2^5-1)-1/2^5+......1/(2^n-1)-1/2^n))-1/2(1/(2^n-1)-1/2^n))即t<1/2(1/(2^4-1)-1/2^4)+t/2-1/2(1/(2^n-1)-1/2^n))<1/2(1/(2^4-1)-1/2^4)+t/2，得t<(1/(2^4-1)-1/2^4)=1/(15x16)=1/240。所以，1/a1+1/a2+...1/an<1/2+1/20+1/56+1/272+1/240<7/12。

追问

高手啊你怎么想的

追答

这个运算量大，就是反复是试出来的。

Answer 2

转证：1/a1+1/a2+1/a3+……+1/an<7/12 - 1 / (2^(n+3)+1)
归纳法 n = 1,2,3,4自己算~~
n足够大时，证n+1，只需证
1/ a(n+1) < 1 / (2^(n+3)+1) - 1 / (2^(n+4)+1)
右边 = 2^(n+3) /(2^(n+3)+1)/(2^(n+4)+1) = 1 / (2^(-n-3)+1)/(2^(n+4)+1)
> 1 / (1+1)/(2^(n+4)+1) = 1 / (2^(n+5)+2)
左边 = 1/ (4^n +(-2)^n) < 1/(4^n - 2^n)

显然当 n足够大时，4^n - 2^n > 2^(n+5)+2, 所以左边<右边
由归纳法得证