[tex]\it n\in\mathbb{Z},\ \dfrac{2n+4}{n^2+1}\in\mathbb{Z}\ \Rightarrow\ \begin{cases}\it\ n^2+1\leq2n+4\ \ \ \ \ (1)\\ \\ \it n^2+1|2n+4\ \ \ \ \ (2)\end{cases}[/tex]
[tex]\it (1) \Rightarrow n^2-2n+1\leq4 \Rightarrow (n-1)^2\leq4 \Rightarrow \sqrt{(n-1)^2}\leq\sqrt4 \Rightarrow \\ \\ |n-1| \leq2 \Rightarrow -2\leq n-1\leq2|_{+1} \Rightarrow -1\leq n\leq 3 \Rightarrow n\in\{-1,\ 0,\ 1,\ 2,\ 3\}\ \ \ (3)\\ \\ (2),\ (3) \Rightarrow n\in\{-1,\ 0,\ 1,\ 3\}[/tex]
_____________________________________
[tex]\it a=\dfrac{1}{10} + \dfrac{1}{11}+ \dfrac{1}{12}+\ ...\ + \dfrac{1}{20} = (\dfrac{1}{10}+ \dfrac{1}{20} ) +( \dfrac{1}{11}+ \dfrac{1}{12}+\ ...\ + \dfrac{1}{19} ) <\\ \\ \\ < \dfrac{3}{20} + \underbrace{\dfrac{1}{11}+ \dfrac{1}{11}+\ ...\ + \dfrac{1}{11} }_{9\ termeni} \Rightarrow a< \dfrac{3}{20} + \dfrac{9}{11} \Rightarrow a< \dfrac{213}{220} <1\Rightarrow a\notin\mathbb{N}[/tex]
