Question

求证：log以n为底(n+1)>log以(n+1)为底(n+2)

Answer 1

证：logn(n+1)=ln(n+1)/ln(n)=1+ln[(n+1)/n]/ln(n)

log(n+1)(n+2)=1+ln[(n+2)/(n+1)]/ln(n+1)

∵(n+1)/n>(n+2)/(n+1)

∴ln[(n+1)/n]>ln[(n+2)/(n+1)]

又ln(n)log(n+1)(n+2)