若n是正整数,且n>1,则logn(n+1)与log(n+1)(n+2)的大小关系为?
因为log(n+1)[n+2]+log(n+1)n=log(n+1)(n^2+2n)=2 所以 log(n+1)(n+2)<logn(n+1) (这是因为它们加同一个数 log(n+1)n 后,一个比2大,一个比2不小)
log(n+1)(n+2)-log(n)(n+1) (换底,使底数为n+1,并且予以省略) =log(n+2)-1/logn =[lognlog(n+2)-1]/logn 因为lognlog(n+2)0 故log(n+2)log(n+1)(n+2).
a=logn(n+1),n^a=n+1 b=log(n+1)(n+2),(n+1)^b=n+2, n^a+1=(n+1)^b>n^b+1^b=n^b+1 n^a>n^b,a>b logn(n+1)>log(n+1)(n+2)
loga[b]中a为底数。
1。logn[n+1]=1/log(n+1)[n],
2。log(n+1)[n+2]=
=log(n+1)[n+1]+log(n+1)[(n+2)/(n+1)]=
=1+log(n+1)[(n+2)/(n+1)]≤
≤1+log(n+1)[(n+1)/n]
3。
{log(n+1)[n+2]}/{logn[n+1]}=
={log(n+1)[n+2]}{log(n+1)[n]}=
={log(n+1)[n+2]}{log(n+1)[n+1]+log(n+1)[n/(n+1)]}=
={log(n+1)[n+2]}{1-log(n+1)[(n+1)/n]}≤
≤{1+log(n+1)[(n+1)/n]}{1-log(n+1)[(n+1)/n]}=
=1-{log(n+1)[(n+1)/n]}^2
log(n+1)[n+2]<logn[n+1]。
。
解: n是正整数且n>1 所以 n(n+1)〈(n+1)(n+2) 一,当log底数a范围 01的时候 logn(n+1)〈log(n+1)(n+2) 三,挡当log底数a范围 a=1得时候,不成立 因为1得任何次方都是1