`sum_(n=1)^oo (4n+1)/(3n-1)` Determine the convergence or divergence of the series.

`sum_(n=1)^oo(4n+1)/(3n-1)`

For the given series `a_n=(4n+1)/(3n-1)`

`a_n=(4+1/n)/(3-1/n)`

`lim_(n->oo)a_n=lim_(n->oo)(4+1/n)/(3-1/n)`

`=4/3!=0`

As per the n'th term test for divergence,

If `lim_(n->oo)a_n!=0` , then `sum_(n=1)^ooa_n` diverges,

So, for the given series `lim_(n->oo)a_n=4/3!=0`

Hence the series diverges.

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