We want to check if the following sequence converges (absolutely):

$$\sum\_{k=1}^n (-1)^k \frac{3^k+2}{4^k -6}$$

Since we have $(-1)^k$, we can use the Leibniz criterion.

We have to prove that $a\_n$ is a null sequence and that it is monotonically decreasing.

$$\lim\_{k \to \infty} \frac{3^k+2}{4^k-6} = \lim\_{k \to \infty} \frac{(\frac{3}{4})^k + \frac{2}{4^k}}{1 - \frac{6}{4^k}} = 0$$

To prove that it is monotonous decreasing, we have to show $a\_n \geq a\_{n+1}$

$$a\_n \geq a\_{n+1} \\ \Leftrightarrow \frac{3^k+2}{4^k-6} \geq \frac{3^{k+1} +2}{4^{k+1}-6} $$

But how do we continue from here?