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:
$$\sum_{n=1}^{\infty} \left(\frac{a n+2}{3n+1}\right)^{n}$$
 A: From my experience, whenever you have an entire expression raised to the $n^\text{th}$ power, first try using the $n$-th Root Test. As a reminder, the $n$-th Root Test states the following:
Let $r = \lim\limits_{n \rightarrow \infty}(a_n)^{1/n}$. Then three possibilities exist:


*

*$r < 1 \implies $ the series converges

*$r = 1 \implies $ inconclusive, requires further consideration

*$r > 1 \implies $ the series diverges


This is not the entire answer, but I will say that, if you use this Root Test, each of the three cases above will give you a range of values that $a$ is allowed to be. If you need further hints, just ask. (In the future, though, be sure to state what you're stuck on; vague questions usually elicit vague--or no--answers.)
Good luck!!
A: 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.
