Question

已知数列cn=2n^2+3n+1，求证：当n>=2时，数列{1/cn}的前n项和小于5/12

Answer 1

1/cn=(2n+1)(n+1)
数列{1/cn}的前n项和=1/(2*3)+1/(3*5)+1/(4*7)+......+1/(2n+1)(n+1)
1/(3*5)+1/(4*7)+......+1/(2n+1)(n+1)<1/2(2*3)+1/2(3*4)+.......+1/2n(n+1)
1/2(2*3)+1/2(3*4)+.......+1/2n(n+1)=1/2(1/2-1/3+1/3-1/4+...+1/n-1/n+1)=1/2(1/2-1/(n+1))<1/4
数列{1/cn}的前n项和<1/6+1/4=5/12

Answer 2

cn=2n^2+3n+1,Cn=(2n+1)(n+1) 1/cn=(1/n+1-1/2n+1)/n＜1/2（1/n-1/(n+1)）
{1/cn}的前n项和＜1/6+1/2（1/2-1/(n+1)）＜5/12

Answer 3

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