证明:
由题意可知:原式即证:1/3+1/6+1/11+…+1/(2^n+n)<3/4 n=1,2,3…
n=1时,1/3<3/4,成立;
同理n=2,3时,原式都成立;
n>=4时,因为显然有1/(2^n+n)<1/2^n
1/3+1/6+1/11+1/20+1/37+…+1/(2^n+n)<1/3+1/6+1/11+1/16+1/32+…+1/2^n
1/3+1/6+1/11+1/16+1/32+…+1/2^n
=(1/3+1/6+1/11)+1/8-1/2^n<3/4
所以n属于N*,n>=4时,原式都成立
得证