A sequence $g_t (t=0,\cdots,\infty$) is such that the relation between successive terms is given implicitly by:
$$\frac{1+g_t}{1+g*}=\frac{1+cg_t}{1+cg_{t+1}}\qquad(1)$$
The initial value $g_0$ is not given but may be taken to be such that $|g_0|<1$. The $g_t$'s, $g*$ and $c$ may be positive, zero or negative.
Question What is the condition on $c$ for $g_t$ to converge to $g*$?
Motivation The question is considered in an old economics book (A), where $g_t$ is the growth rate (variable) of capital, $g*$ is the growth rate (assumed constant) of labour and $c$ is derived from production coefficients. The author manipulates $(1)$ to obtain the following:
$$\frac{c(1+g*)}{1+cg_{t+1}}=\frac{c(1+g_t)}{1+cg_t}=1+\frac{c-1}{1+cg_t}$$
$$\frac{c(1+g*)}{1+cg_{t+1}}-\frac{c-1}{1+cg_t}=1=\frac{c(1+g*)}{1+cg*}-\frac{c-1}{1+cg*}$$
I can follow the algebra but cannot understand why the author concludes from the above that "$g_t$ converges to $g*$ if $\frac{c-1}{c(1+g*)}$ is less than $1$ in absolute value".
My Attempt Starting from $(1)$ I reasoned informally as follows. If $-1 < c < 1$, then the right-hand side of $(1)$ is closer to $1$ than $\frac{1 + g_t}{1 + g_{t+1}}$. Hence the left-hand side, $\frac{1 + g_t}{1 + g*}$ is also closer to $1$ than $\frac{1 + g_t}{1 + g_{t+1}}$. These fractions have the same numerators, $1+g_t$, so the denominator of the former must be further from $1 + g_t$ than the denominator of the latter. Hence $1 + g_{t+1}$ must lie between $1 + g_t$ and $1 + g*$, so that $g_{t+1}$ lies between $g_t$ and $g*$, implying that $g_t$ converges to $g*$.
However, my conclusion appears inconsistent with the author's. Mine implies that there is convergence if $c$ is very close to zero, whereas the author's formula $|\frac{c-1}{c(1+g*)}|$ will exceed $1$ for sufficiently small $c$.
Reference
(A) Hicks, J (1965) Capital and Growth Oxford University Press pp 186-7
Update 8/6/2020
I can see now that the last step in my attempt is fallacious: $g_{t+1}$ lies between $g_t$ and $g*$ for all $t$ does not imply that $g_t$ converges to $g*$.
$$\frac{c(1+g*)}{1+cg_{t+1}}-\frac{c-1}{1+cg_t}=\frac{c(1+g*)}{1+cg*}-\frac{c-1}{1+cg*}$$ can be written as $$\frac{1}{1+cg_{t+1}}-\frac{1}{1+cg*}=\frac{c-1}{c(1+g*)}\bigg(\frac{1}{1+cg_t}-\frac{1}{1+cg*}\bigg)$$ which is equivalent to $$\frac{g*-g_{t+1}}{(1+cg_{t+1})(1+cg*)}=\frac{c-1}{c(1+g*)}\cdot \frac{g*-g_t}{(1+cg_t)(1+cg*)}$$ from which we have $$\frac{g*-g_{t}}{(1+cg_{t})(1+cg*)}=\bigg(\frac{c-1}{c(1+g*)}\bigg)^t\frac{g*-g_0}{(1+cg_0)(1+cg*)}$$
So, we see that if $\bigg|\frac{c-1}{c(1+g*)}\bigg|\lt 1$, then $g_t$ converges to $g*$.