[tex]\text{A} = 1 + \dfrac{1}{2} + \dfrac{1}{3} +\dfrac{1}{4}+ ... + \dfrac{1}{2017} + \\ \\ +\dfrac{1}{2} + \dfrac{2}{3} +\dfrac{3}{4} +... + \dfrac{2016}{2017} =\\ \\ =\sum\limits_{k=1}^{2017} \dfrac{1}{k} + \sum\limits_{k=1}^{2017} \dfrac{k-1}{k}= \\ =\sum\limits_{k=1} ^{2017}\left(\dfrac{1}{k} + \dfrac{k-1}{k}\right)= \\ = \sum\limits_{k=1}^{2017}\left(\dfrac{1}{k} +\dfrac{k}{k}- \dfrac{1}{k} \right) = \\ =\sum\limits_{k=1}^{2017}\left(\dfrac{1}{k} + 1-\dfrac{1}{k}\right) = \\ = \sum\limits_{k=1}^{2017} \left(\dfrac{1}{k} - \dfrac{1}{k}+1\right) = \\ =\sum\limits_{k=1}^{2017} 1 = \\ \\= 2017 [/tex]
