Imagine that $x \in (0,1]$ and $f$ is a continuous function such that $0 <f(x) < x$. Also, for any $k \in \mathbf N$ define $f^{k}(x)$ recursively by $f^{k} (x) = f(f^{k-1} (x))$.
Are there any theorems that put simple, but also fairly general conditions on $f(x)$ that ensures that the series $f(x)+ f^{2} (x) + f^{3}(x)+ …$ is convergent? Thanks.
A condition for the convergence of the sequence $\{f^k(x)\}$ is that $f$ is a contraction: if there exists a number $0 \le \theta < 1$ with the property that $|f(x) - f(y)| \le \theta |x-y|$ for all $x$ and $y$, it is fairly routine to show that $\{f^k(x)\}$ is a Cauchy sequence for any $x$ in the domain of $f$, and moreover that the limit is the same for all $x$.
This suggests an analogy for series.
If there is a constant $0 \le \theta < 1$ with the property that $0 < f(x) \le \theta x$ for all $x$, then $f: (0,1] \to (0,\theta]$ and by iterating the inequality you find $0 \le f^k(x) \le \theta^{k-1}x \le \theta^{k-1}$ for all $x$. The series is then seen to be convergent using the comparison test with a convergent geometric series.