Răspuns:
7/6
Continuarea din imagine:
[tex]\lim\limits_{n\to \infty}\Big[2\Big(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\Big) - \\ -3\Big(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\Big)\Big]= \\ \\ = 2\Big(1+\dfrac{1}{2}+\dfrac{1}{3}\Big)-3\Big(\dfrac{1}{2}+\dfrac{1}{3}\Big) = \\ \\ = 2+1+\dfrac{2}{3}-\dfrac{3}{2}-1 = \dfrac{12+4-9}{6} = \boxed{\dfrac{7}{6}}[/tex]
