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We want to judge whether the system is stable or not.

Given the below transfer function.

$$ H\left( z \right) =\frac{\left( 1+2 z^{-1} \right) }{\left( 2+z^{-1} \right) } $$

$$ H\left( z \right) =2-\frac{3}{2+z^{-1} } $$

$$ 2+\frac{-3}{2+z^{-1} } $$

$$S:=\frac{-3}{2+z^{-1} }$$

$$ S ~~\text{is the sum of each element of the geometric sequence.} $$

$$ -3 ~~ \leftarrow~~ \text{initial term} $$

$$ z^{-1} ~~ \leftarrow~~ \text{common ratio} $$

$$ i \geq1 \rightarrow \text{ith term} = -3 \cdot \left( z^{-1} \right) ^{i-1} $$

$$ = -3 \cdot z^{1-i} =-3 \cdot z^{-0}~,~-3z^{-1}~,~-3 z^{-2} ~,~ \cdot\cdot\cdot $$

$$ H\left( z^{} \right) =\sum_{ n=-\infty }^{ \infty } h\left[ n \right] z^{-n} $$

$$ \sum_{ n=-\infty }^{ \infty } \left( 2 \delta\left[n \right] +\left( -3 \right) u\left[ n \right] \right)z^{-n} $$

$$\displaystyle \therefore ~~ h\left[ n \right] =2 \delta\left[n \right] +\left( -3 \right) u\left[ n \right] $$

$$ \sum_{ n=-\infty }^{ \infty } \left| h\left[ n \right] \right| $$

$$ = \sum_{ n=-\infty }^{ \infty } \left| 2 \delta\left[n \right] +\left( -3 \right) u\left[ n \right]\right| = \infty $$

$$ \therefore ~~ ~~\text{The system is unstable.}~~ $$

However from the another approach,

$$ H\left( z \right) =2-\frac{3}{2+z^{-1} } $$

$$ = \frac{2 \left( 2+ z^{-1} \right) -3}{2+z^{-1} } $$

$$2+z^{-1} = 0 $$

$$ 2+\frac{1}{z^{} } =0 $$

$$ \frac{1}{z^{} } =-2 $$

$$ z=-\frac{1}{2} =-0.5 ~~ \leftarrow~~ ~~\text{pole}~~ $$

$$ ~~\text{Since }~~ \left| \text{pole} \right| =\left| -0.5 \right| =\left| 0.5 \right| <1 ~~\text{is held, the system is stable.}~~ $$

Why the different concolusions were gained?

What I've been missing?

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The system is stable. Your mistake is in the application of the formula for the geometric series:

$$\begin{align}-\frac{3}{2+z^{-1}}&=-\frac32\frac{1}{1+\frac12z^{-1}}\\&=-\frac32\sum_{n=0}^{\infty}\left(-\frac12\right)^nz^{-n}\end{align}$$

Hence, the system's impulse response is

$$h[n]=2\delta[n]-\frac32\left(-\frac12\right)^nu[n]$$

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