已知数列{an}满足an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) 1.数列{an}是递增数列还是递减数列2.证明 an≥1/2对一切正整数恒成立
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![已知数列{an}满足an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) 1.数列{an}是递增数列还是递减数列2.证明 an≥1/2对一切正整数恒成立](/uploads/image/z/13390434-18-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3an%3D1%2F%28n%2B1%29%2B1%2F%28n%2B2%29%2B1%2F%28n%2B3%29%2B...%2B1%2F%282n%29+1.%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%98%AF%E9%80%92%E5%A2%9E%E6%95%B0%E5%88%97%E8%BF%98%E6%98%AF%E9%80%92%E5%87%8F%E6%95%B0%E5%88%972.%E8%AF%81%E6%98%8E+an%E2%89%A51%2F2%E5%AF%B9%E4%B8%80%E5%88%87%E6%AD%A3%E6%95%B4%E6%95%B0%E6%81%92%E6%88%90%E7%AB%8B)
已知数列{an}满足an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) 1.数列{an}是递增数列还是递减数列2.证明 an≥1/2对一切正整数恒成立
已知数列{an}满足an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) 1.数列{an}是递增数列还是递减数列
2.证明 an≥1/2对一切正整数恒成立
已知数列{an}满足an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) 1.数列{an}是递增数列还是递减数列2.证明 an≥1/2对一切正整数恒成立
1.
a(n+1)-an
=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[(n+1)+(n-1)]+1/[(n+1)+n]+1/[(n+1)+(n+1)]
-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(2n+1)+1/(2n+2) -1/(n+1)
=1/(2n+1)-1/(2n+2)>0
a(n+1)>an,数列是递增数列.
2.
证:
n=1时,a1=1/(1+1)=1/2,不等式成立.
由第1问得数列是递增数列,即n≥2时,an>a1 an>1/2,不等式成立.
综上,得an≥1/2对一切正整数n恒成立.
解
1、数列{an}是递增数列
2、因为a1=1/(1+1)=1/2
又因为{an}是递增数列
所以an≥1/2对一切正整数恒成立
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