Croatian Mathematical Competitions

Let us denote the sum by $S$:
$$S = \log_2 \left(1 + \frac{1}{1}\right) + \log_2 \left(1 + \frac{1}{2}\right) + \dots + \log_2 \left(1 + \frac{1}{2014}\right)$$

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:
$$S = \log_2 2 - \log_2 1 + \log_2 3 - \log_2 2 + \log_2 4 - \log_2 3 + \dots + \log_2 2015 - \log_2 2014$$

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,
$$\log_2 \left(1 + \frac{1}{1}\right) + \log_2 \left(1 + \frac{1}{2}\right) + \dots + \log_2 \left(1 + \frac{1}{2014}\right) < 11.$$