2. Zadatak

Prvo rješenje.

Vrijedi

$$\log_2\left(1+\frac1k\right) =\log_2\left(\frac{k+1}{k}\right) =\log_2(k+1)-\log_2(k).$$

Zbrajanjem takvih izraza za sve $k\in\{1,2,\ldots,2014\}$ imamo

$$\begin{aligned} &\log_2\left(1+\frac11\right)+\log_2\left(1+\frac12\right)+\log_2\left(1+\frac13\right)+\cdots+\log_2\left(1+\frac1{2014}\right)\\ &=(\log_2 2-\log_2 1)+(\log_2 3-\log_2 2)+(\log_2 4-\log_2 3)+\cdots+(\log_2 2015-\log_2 2014)\\ &=\log_2 2015-\log_2 1=\log_2 2015. \end{aligned}$$

Budući da je $2015<2048=2^{11}$ slijedi $\log_2(2015)<11$, pa vrijedi i tvrdnja zadatka.

Drugo rješenje.

Zbog svojstava logaritama je

$$\begin{aligned} &\log_2\left(1+\frac11\right)+\log_2\left(1+\frac12\right)+\log_2\left(1+\frac13\right)+\cdots+\log_2\left(1+\frac1{2014}\right)\\ &=\log_2\left(\left(1+\frac11\right)\left(1+\frac12\right)\left(1+\frac13\right)\cdots\left(1+\frac1{2014}\right)\right)\\ &=\log_2\left(\frac21\cdot\frac32\cdot\frac43\cdots\frac{2015}{2014}\right)=\log_2(2015). \end{aligned}$$

Budući da je $2015<2048=2^{11}$ slijedi $\log_2(2015)<11$, pa vrijedi i tvrdnja zadatka.