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 $1/1999<\ln(1999/1998)$.

Thursday, December 1st

Problem
Find
$S=\sum_{i=0}^{101} \frac{x_i^3}{1-3x_i+3x_i^2}$
where $x_i=i/101$.


Solution
Since
$\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{(1-x_i)^3+x_i^3}$

is symmetric when exchanging $x_i$ by $1-x_i$, we have that every such pair will contribute to the sum by one, and there are 51 such pairs, hence $S=51$.