Use a logarithmic fixed-point graph to compare terms

Solving $\ln(x+2)-x/5=0$ numerically gives $\alpha\approx13.80007271$, or $13.8$ to 3 significant figures. With $h(x)=\ln(x+2)-x/5$, $h(0)>0$, $h^\prime(x)=1/(x+2)-1/5$ changes sign at $3$, and $h(x)\to-\infty$. Thus there is exactly one positive root.

If the positive sequence converges to $\ell\ge0$, then $\ell=5\ln(\ell+2)$, so $h(\ell)=0$. Since $h(0)>0$ and the positive root is unique, $\ell=\alpha$.

For $x_1=1$, the printed selected terms are:

The terms increase towards $\alpha$. For $x_1=20$:

The terms decrease towards $\alpha$. Rounded repetitions of 13.8 are not exact equality.

The calculator screenshots also show all eleven rows at their displayed precision. Enter the same recurrence in sequence mode and change only the initial value between the two runs:

The shorter printed tables above use different rounding from the calculator screenshots. Both display versions are retained; these rounded repetitions do not mean the recurrence has reached its exact fixed point.

$x_{n+1}-x_n=5h(x_n)$. For $0<x_n<\alpha$, $h(x_n)>0$, giving $x_{n+1}>x_n$; for $x_n>\alpha$, $h(x_n)<0$, giving $x_{n+1}<x_n$. Since $5\ln(x+2)$ is increasing and fixes $\alpha$, a term below $\alpha$ stays below it, and a term above stays above. The two observed sequences are therefore monotone and bounded and converge to $\alpha$.