Question

Calcular o limite de an e determinar se a sequencia converge ou diverge.

Answer 1

Para verivficar se uma sequência converge ou diverge, precisamos calcular o limite do termo geral $a_n,$ quando $n$ tende a infinito (caso exista):

$\underset{n\to \infty}{\mathrm{\ell im}}\;a_n\\\\ =\underset{n\to \infty}{\mathrm{\ell im}}\;\dfrac{4n^4+3n^2}{n+2n^3}\\\\\\ =\underset{n\to \infty}{\mathrm{\ell im}}\;\dfrac{n^4\cdot \left(4+\frac{3}{n^2}\right)}{n^3\cdot \left(\frac{1}{n^2}+2 \right )}\\\\\\ =\underset{n\to \infty}{\mathrm{\ell im}}\;n\cdot \dfrac{4+\frac{3}{n^2}}{\frac{1}{n^2}+2}\\\\\\ =\underset{n\to \infty}{\mathrm{\ell im}}\;b_n\cdot c_n~~~~~~\mathbf{(i)}$


onde

$b_n=n~~\text{ e }~~c_n=\dfrac{4+\frac{3}{n^2}}{\frac{1}{n^2}+2}$

_________________

Sabemos que

$\bullet\;\;\underset{n\to \infty}{\mathrm{\ell im}}\;b_n\\\\ =\underset{n\to \infty}{\mathrm{\ell im}}\;n\\\\ \infty$


$\bullet\;\;\underset{n\to \infty}{\mathrm{\ell im}}\;c_n\\\\\\ =\underset{n\to \infty}{\mathrm{\ell im}}\;\dfrac{4+\frac{3}{n^2}}{\frac{1}{n^2}+2}\\\\\\ =\dfrac{4+0}{0+2}\\\\\\ =2>0$


Portanto, o limite $\mathbf{(i)}$ é

$\underset{n\to \infty}{\mathrm{\ell im}}\;b_n\cdot c_n=\infty\\\\\\ \therefore~~\boxed{\begin{array}{c}\underset{n\to \infty}{\mathrm{\ell im}}\;\dfrac{4n^4+3n^2}{n+2n^3}=\infty \end{array}}$

A sequência diverge.


Bons estudos! :-)