Answer:
The series diverges
Step-by-step explanation:
First, let's find the sequence:
[tex]4, 8, 12, 16, 20,...,[/tex]
As we can see, they are multiples of 4, so one possible sequence is:
[tex]4,8,12,16,20,...,=4n\hspace{8};\hspace{3}n\in Z[/tex]
Hence, we can represent the series as:
[tex]$\sum_{n=1}^{\infty} 4n[/tex]
Let's use the zero test which states:
[tex]$\Sigma a_n$\\Diverges\hspace{3}if\\\\ \lim_{n \to \infty} a_n \neq 0[/tex]
So:
[tex]\lim_{n \to \infty} 4n= 4 \lim_{n \to \infty} n\\ \\ \lim_{n \to \infty} n=\infty\\ \\4\infty= \infty\\\\Hence\\\\ \lim_{n \to \infty} 4n=\infty[/tex]
Therefore, by the zero test, the series diverges