(Absolute) convergence of $\sum_{k=1}^n (-1)^k \frac{3^k+2}{4^k -6}$ 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?

As Lionel Ricci's comment indicates, rather than trying to use the Leibniz criterion (note that, as described in Leibniz Criterion, this criterion (also called the alternating series test) only indicates the series converges, not that it necessarily converges absolutely), instead use that, for $k \ge 2$, we have $4^k - 6 \gt 4^{k-1}$ and $3^k + 2 \lt 3^{k+1}$. Thus,
$$\frac{3^k+2}{4^k-6} \lt \frac{3^{k+1}}{4^{k-1}} = \frac{3^2(3^{k-1})}{4^{k-1}} = 3^2\left(\frac{3}{4}\right)^{k-1} \tag{1}\label{eq1A}$$
Therefore,
$$\sum_{k=1}^{\infty}\left|(-1)^k\left(\frac{3^k+2}{4^k-6}\right)\right| \lt \frac{5}{2} + \sum_{k=2}^{\infty}9\left(\frac{3}{4}\right)^{k-1} = \frac{5}{2} + \frac{9\left(\frac{3}{4}\right)}{1 - \frac{3}{4}} = \frac{59}{2} \tag{2}\label{eq2A}$$
