1. Zadatak

Za svaki $n\in\{1,2,\ldots,2019\}$ imamo

$$\begin{aligned} |a_n-a_{n+1}| &=\left|a_n-a_n\left(1+\frac{i}{\sqrt{n+1}}\right)\right|\\ &=|a_n|\cdot\left|\frac{i}{\sqrt{n+1}}\right| =\frac{|a_n|}{\sqrt{n+1}}. \end{aligned}$$

Nadalje je

$$\begin{aligned} |a_n| &=|1+i|\cdot\left|1+\frac{i}{\sqrt2}\right| \cdot\left|1+\frac{i}{\sqrt3}\right|\cdots\left|1+\frac{i}{\sqrt n}\right|\\ &=\sqrt{1+1}\cdot\sqrt{1+\frac12}\cdot\sqrt{1+\frac13}\cdots\sqrt{1+\frac1n}\\ &=\sqrt2\cdot\frac{\sqrt3}{\sqrt2}\cdot\frac{\sqrt4}{\sqrt3}\cdots\frac{\sqrt{n+1}}{\sqrt n}\\ &=\sqrt{n+1}. \end{aligned}$$

Zato je $|a_n-a_{n+1}|=1$ za sve $n\in\{1,2,\ldots,2019\}$.

Konačno je

$$|a_1-a_2|+|a_2-a_3|+\cdots+|a_{2019}-a_{2020}|=2019\cdot1=2019.$$