Mathematica competitions in Croatia

Let us denote the sum by $S$: $$S={\log }_2\left(1+\frac{1}{1}\right)+{\log }_2\left(1+\frac{1}{2}\right)+\cdots +{\log }_2\left(1+\frac{1}{2014}\right)$$

We have: $${\log }_2\left(1+\frac{1}{k}\right)={\log }_2\left(\frac{k+1}{k}\right)={\log }_2(k+1)-{\log }_{2}k$$

Therefore, $$S=\sum _{k=1}^{2014}\left[{\log }_2(k+1)-{\log }_{2}k\right]$$

This is a telescoping sum: $$S={\log }_{2}2-{\log }_{2}1+{\log }_{2}3-{\log }_{2}2+{\log }_{2}4-{\log }_{2}3+\cdots +{\log }_{2}2015-{\log }_{2}2014$$

All terms except the first ${\log }_{2}2$ and the last $-{\log }_{2}2014$ cancel, so: $$S={\log }_2(2015)-{\log }_{2}1={\log }_2(2015)$$

We need to show that ${\log }_2(2015)<11$.

But $2^{11}=2048>2015$, so ${\log }_2(2015)<11$.

Therefore, $${\log }_2\left(1+\frac{1}{1}\right)+{\log }_2\left(1+\frac{1}{2}\right)+\cdots +{\log }_2\left(1+\frac{1}{2014}\right)<11.$$