Uniform convergence of $\sum f(x)^n$ 
Let $f:X\to\mathbb{R}$ be such that $\sup\{|f(x)|:x\in X\}<1.$ Show that $\sum_{n=1}^{\infty} f(x)^n$ converges and compute the sum..

Every value given by $f$ is less than one, then if $b=f(x)$ is $b<\frac{p}{q}$ with $p<q$; the limit $\left(\frac{p}{q}\right)^n\to 0$ if $n\to +\infty$, then $f(x)\to 0 \;\forall x\in X$.Now how can I show the convergence of the sum?, I thought I could use $f(x)^n\geq f(x)^{n+1}$ to bound the sum but there is no guarantee that $\sum_if^k(x)$ will converge (actually, it doesn't converge).
Another approach could be use that $f(x)\in [0,1)$, then exists an $(1/n^2)\in[0,1)$ such that $f(x)<1/n^2$, except for a finitely many terms; letting those terms outside and using $1/n^2$ as a boundary for $f$ I conclude that $\sum_n f(x)^n$ converges. This doesn't solve the other question, why the convergence is uniform and what is exactly the value of the sum?
 A: Hints:
Let $a_n=f^n(x)$. So can you proceed from here $\sqrt[n]{a_n}=f(x)<1$?
Moreover, do you know that $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ if $x\in (-1,1)$?
A: Uniform convergence of $\sum_1^\infty (f(x))^n$ indeed follows from the given hypotheses;  I reason as follows:
We are given that $\sup \{ \vert f(x) \vert : x \in X \} < 1$; set  $\delta = \sup \{ \vert f(x) \vert : x \in M \}$; then $\delta < 1$ and $\vert f(x) \vert \le \delta$ for all $x \in X$.  We assume $\delta > 0$ lest the problem be trivialized.  Let $S_m(x)$ denote the sequence of partial sums of the series $\sum_1^\infty (f(x))^n$; then
$S_m(x) = \sum_1^m (f(x))^n.  \tag{1}$
Choose positive integers $k, l$ with $k > l$; then
$S_k(x) - S_l(x) = \sum_{l + 1}^k (f(x))^n = (f(x))^l \sum_1^{k -l} (f(x))^n, \tag{2}$
whence
$\vert S_k(x) - S_l(x) \vert = \vert (f(x))^l \vert \vert \sum_1^{k - l} (f(x))^n \vert = \vert f(x) \vert^l \vert \sum_1^{l - k} (f(x))^n \vert$
$\le \vert f(x) \vert^l \sum_1^{k - l} \vert f(x) \vert^n \le \delta^l\sum_1^{k - l} \delta^n. \tag{3}$
We examine $\sum_1^{k - l} \delta^n$, noting that $\sum_1^\infty \delta^n$ is a geometric series with ratio $\delta < 1$, and hence it converges to the well-known sum
$\sum_1^\infty \delta^n = \delta \sum_0^\infty \delta^n = \dfrac{\delta}{1 - \delta};\tag{4}$
we clearly have
$\sum_1^{k - l} \delta^n < \sum_1^\infty \delta^n = \dfrac{\delta}{1 - \delta}; \tag{5}$
combining (3) and (5) shows that
$\vert S_k(x) - S_l(x) \vert < \dfrac{\delta^{l + 1}}{1 - \delta} \tag{6}$
holds for all $k > l$, independent of $x \in X$.  Since $\delta > 1$, (6) shows that
$\vert S_k(x) - S_l(x) \vert$ may be made arbitrarily small provided $l$ is sufficiently large.  The sequence $S_m(x)$ is thus Cauchy for all $x \in X$, and hence converges; in fact, the sum is given by the same formula for geometric series (4) as is used to bound $\vert S_k(x) - S_l(x) \vert$ in (6), viz.
$\sum_1^\infty (f(x))^n = f(x) \sum_0^\infty (f(x))^n = \dfrac{f(x)}{1 - f(x)}.  \tag{7}$
(4) and (7) follow from the well known formula for the sum of a finite geometric series
$(1 - a)(\sum_0^m a^i) = 1 - a^{m + 1} \tag{8}$
or
$\sum_0^m a^i = \dfrac{ 1 - a^{m + 1}}{1 - a}; \tag{9}$
when $\vert a \vert < 1$, taking $m \to \infty$ in (8) or (9) yields the result that
$\sum_0^m a^i = \dfrac{1}{1 - a}, \tag{10}$
vindicating both (4) and (7).  Finally, we see that the convergence of the sequence $S_n(x)$ is in fact uniform on $X$, again via (6); by taking $l$ sufficiently large, $\vert S_k(x) - S_l(x) \vert$ may be made as small as we please, globally on $X$, without reference to any particular $x \in X$.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
