Convergence of a series with a parameter $a>0$ Hello I have this exercise:
Please help me determine what values ​​of $a > 0$ the series below converges:
$$\sum_{n=1}^{\infty} \left(\frac{a n+2}{3n+1}\right)^{n}$$

Hint:
When $a=3$,
$$\sum_{n=1}^{\infty} \left(\frac{3 n+2}{3n+1}\right)^{n}
=\sum_{n=1}^{\infty} \left(1+\frac1{3n+1}\right)^{n}
=\sum_{m=4,\text{ step }3}^{\infty} \left(1+\frac1m\right)^{(m-1)/3}\\
=\sum_{m=4,\text{ step }3}^{\infty}{\frac{\sqrt[3]{\left(1+\frac1m\right)^m}}{\sqrt[3]{1+\frac1m}}}.$$
The expression of the numerator should ring a bell.
