Question

Calculati limita:
$\lim_{n \to \infty} a_n$, $a_{n}=$∑ de la k=1 pana la n din $\dfrac{5n^{2}+6k^{2} }{3n^{3}+n+k }$

Answer 1

$\it \dfrac{5n^2+6k^2}{3n^3+n+k} < \dfrac{5n^2+6k^2}{3n^3} < \dfrac{11n^2}{3n^3} \Rightarrow a_n < n\cdot\dfrac{11n^2}{3n^3}=\dfrac{11n^3}{3n^3}\ \longrightarrow\ \dfrac{11}{3}$