How to prove this theorem about the Z transform and final value theorem?
Let me show you a simple way to see this property. Assume $x[k]$ is a causal sequence and let

$$x[\infty]=\lim_{k\rightarrow\infty}x[k]$$

be finite. Then the sequence $x[k]$ can be written as

$$x[k]=x[\infty]u[k]+y[k]\tag{1}$$

where $u[k]$ is the unit step sequence, and $y[k]$ is a causal sequence that decays to zero as $k\rightarrow\infty$. Taking the $\mathcal{Z}$-transform of (1) gives

$$X(z)=x[\infty]\cdot\frac{z}{z-1}+Y(z)\tag{2}$$

The poles of $X(z)$ are determined by the two terms on the right-hand side of (2). Since $y[k]$ is a causal decaying sequence, its $\mathcal{Z}$-transform $Y(z)$ must have all its poles inside the unit circle. Since the first term on the right-hand side of (2) can only contribute a single pole at $z=1$, $X(z)$ cannot have any poles outside the unit circle. If $x[\infty]\neq 0$, $X(z)$ has exactly one pole at $z=1$ from the first term on the right-hand side of (2). If $x[\infty]=0$, i.e. $x[k]$ decays to zero as $k\rightarrow\infty$, then that first term disappears and $X(z)$ has no pole at $z=1$. This verifies the claim: if $x[k]$ has a finite limit as $k\rightarrow\infty$, $X(z)$ cannot have any poles outside the unit circle, and, if $x[\infty]\neq 0$, it has a single pole at $z=1$ in addition to its poles inside the unit circle.

Also note that the final value theorem directly follows from (2). Multiplying both sides of (2) by $z-1$ gives

$$(z-1)X(z)=x[\infty]z+(z-1)Y(z)\tag{3}$$

Taking the limit $z\rightarrow 1$ yields the final value theorem of the $\mathcal{Z}$-transform:

$$\lim_{z\rightarrow 1}(z-1)X(z)=x[\infty]\tag{4}$$

Note that the limit $\lim_{k\rightarrow\infty}x[k]$ must be guaranteed to exist in order for (4) to make any sense.