I tried two approaches and gained the different conclusions of judging the stability of the transfer function of the system 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,
$$ H\left( z  \right) =2-\frac{3}{2+z^{-1} }   $$
$$ = \frac{2 \left( 2+ z^{-1}  \right) -3}{2+z^{-1} }   $$
$$2+z^{-1}  = 0  $$
$$  2+\frac{1}{z^{} } =0  $$
$$  \frac{1}{z^{} } =-2  $$
$$  z=-\frac{1}{2} =-0.5 ~~ \leftarrow~~ ~~\text{pole}~~   $$
$$  ~~\text{Since }~~ \left| \text{pole}  \right|  =\left| -0.5 \right| =\left| 0.5 \right| &lt;1 ~~\text{is held, the system is stable.}~~  $$
Why the different concolusions were gained?
What I've been missing?
The system is stable. Your mistake is in the application of the formula for the geometric series:
$$\begin{align}-\frac{3}{2+z^{-1}}&amp;=-\frac32\frac{1}{1+\frac12z^{-1}}\\&amp;=-\frac32\sum_{n=0}^{\infty}\left(-\frac12\right)^nz^{-n}\end{align}$$
Hence, the system's impulse response is
$$h[n]=2\delta[n]-\frac32\left(-\frac12\right)^nu[n]$$