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Sa se calculeze limita sirului an = n⁻³ + n⁻¹

Răspuns :

[tex]a_n= n^{-3}+n^{-1}=\dfrac{1}{n^3}+\dfrac{1}{n}\\\\\displaystyle  \limit\lim_{n\to \infty} a_n=\limit\lim_{n\to\infty}\left(\dfrac{1}{n^3}+\dfrac{1}{n} \right)=0 [/tex]