Given the sum:
[tex]\sum ^{\infty}_{n\mathop=1}\frac{3n^5}{6n^6+1}[/tex]
Analyzing the argument:
[tex]\frac{3n^5}{6n^6+1}=\frac{3n^5}{n^6(6+\frac{1}{n^6})}=\frac{1}{n}\cdot(\frac{3}{6+\frac{1}{n^6}})[/tex]
Where:
[tex]\lim _{n\to\infty}(\frac{3}{6+\frac{1}{n^6}})=\frac{3}{6+0}=\frac{1}{2}\text{ (Bounded)}[/tex]
It is known that the harmonic sum diverges:
[tex]\sum ^{\infty}_{n\mathop=1}\frac{1}{n}=\infty[/tex]
And we have a multiplication between a term that diverges and a bounded term, so we can conclude that the product diverges. Then:
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop=0}\frac{1}{n}\cdot(\frac{3}{6+\frac{1}{n^6}})=\infty \\ \\ \Rightarrow\sum ^{\infty}_{n\mathop{=}1}\frac{3n^5}{6n^6+1}=\infty \end{gathered}[/tex]
