Friday, December 2nd

Problem
Prove that $1/1999<\ln (1999/1998 )<1/1998 $

Solution

Since $1999/1998=1+1/1998$, by the power series expansion of $\ln(1+x)$ we have that $\ln(1999/1998)<1/1998$. Also, since $\ln(1999/1998)=-\ln(1998/1999)=-\ln(1-1/1999)$ and $\ln(1-1/1999)<-1/1999$, we have that $1/1999<\ln(1999/1998)$.