Given the following series:
[tex]\sum\frac{3+2n}{(2+n^2)^2}[/tex]
We will use the limit Comparison Test to determine whether the series is convergent or divergent.
The limit will be as follows:
[tex]\begin{gathered} lim_{n\rightarrow\infty}\frac{a_n}{b_n} \\ a_n=\frac{3+2n}{(2+n^{2})^{2}} \\ b_n=\frac{2n}{n^4}=\frac{2}{n^3} \end{gathered}[/tex]
So, Substitute with (a) and (b) into the limit then calculate the limit:
[tex]\begin{gathered} lim_{n\rightarrow\infty}\frac{3+2n}{(2+n^2)^2}*\frac{n^3}{2} \\ \\ =l\imaginaryI m_{n\operatorname{\rightarrow}\infty}\frac{3n^3+2n^4}{2(4+4n^2+n^4)}=l\imaginaryI m_{n\operatorname{\rightarrow}\infty}\frac{3n^3+2n^4}{8+8n^2+2n^4} \end{gathered}[/tex]
Divide both the numerator and the denominator by n⁴ then substitute
n = ∞, so, the result will be:
[tex]=\frac{2}{2}=1[/tex]
As the value of the limit is finite and positive
So, both the series are convergent or divergent
The series (b) is convergent
So, the given series is convergent
