Answer
Diverges
Work Step by Step
$\lim _{n\rightarrow \infty }\dfrac {4n+1}{3n-1}=\dfrac {4+\dfrac {1}{n}}{3-\dfrac {1}{n}}=\dfrac {4+0}{3-0}=\dfrac {4}{3}\neq 0$
$\Rightarrow \sum ^{\infty }_{n=0}\dfrac {4n+1}{3n-1}$ diverges
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