(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?
 A: 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,
$$\sum_{k=1}^{\infty}\left|(-1)^k\left(\frac{3^k+2}{4^k-6}\right)\right| \lt \frac{5}{2} + \sum_{k=2}^{\infty}9\left(\frac{3}{4}\right)^{k-1} = \frac{5}{2} + \frac{9\left(\frac{3}{4}\right)}{1 - \frac{3}{4}} = \frac{59}{2} \tag{2}\label{eq2A}$$
A: The series $\sum_{k=1}^\infty a_k$ is said to converge absolutely if $\sum_{k=1}^\infty \vert a_k \vert < \infty$, or rather
$$
\lim_{n \to \infty} \sum_{k = 1}^n \vert a_k \vert < \infty.
$$
Notice that the Leibniz test determines whether the series converges, not whether it converges absolutely. A typical example when this test is used is to determine whether the series with $a_k = (-1)^k n^{-1}$ (i.e. the so-called alternating harmonic series) converges or not (it does, but not absolutely!).
To figure out if your series converges or not, one can consider some rather simple estimates. You can check that for all $k \geq 1$ we have $3^k + 2 \leq 2\cdot 3^k$, and for all $k \geq 2$ we have $4^k - 6 \geq (1/2)4^k$. Therefore
$$
\sum_{k=1}^n \left\vert (-1)^k \frac{3^k + 2}{4^k - 6} \right\vert \leq \frac{5}{2} + 4 \cdot \sum_{k=2}^n \left( \frac{3}{4} \right)^k.
$$
What can you say about this rightmost sum?
