Uniform convergence of a series For which $x \ge 0$ does the power series
$$
\sum_{n=0}^\infty \frac x{(1+x)^n}
$$
converge uniformly?

Okay, I see that for $n \ge 2$ we obtain from upon taking a derivative and setting to $0$ the critical value
$$
x=\frac 1{n-1}.
$$
At that critical value,
$$
\frac{x}{(1+x)^n} = \frac{(n-1)^{n-1}}{n^n}.
$$
That value is the maximum over all $x \ge 0$, which means
$$
\left|\frac{x}{(1+x)^n}\right| \le \frac{(n-1)^{n-1}}{n^n}.
$$
for all $x \ge 0$, $n \ge 2$.
Am I now left to establish that
$$
\sum_{n=0}^\infty \frac{(n-1)^{n-1}}{n^n} &lt; \infty,
$$
so that, using the Weierstrass M-test, the power series converges uniformly for all $x \ge 0$?
You need to bound away from the origin as this series is geometric with common ratio $1/(1+x)$, (ignoring the finite x) so you need 
$$
\frac{1}{1+x}&lt;1\Rightarrow 1&lt;1+x\Rightarrow x&gt;0
$$
You can also see this by noting that if $x\in (\epsilon,\infty)$ you can bound 
$$
\frac{1}{1+x}\leq\frac{1}{1+\epsilon}
$$
And then apply the Weirstrass M test.
$$
\sum_{n=0}^\infty \frac x{(1+x)^n} = x\sum_{n=0}^\infty \frac {1}{(1+x)^n}= x\sum_{n=0}^\infty (1+x)^{-n}= x(\frac{1}{x}+1)= (1+x)
$$


  $$
\lim\limits_{n \rightarrow \infty}{(1+x)} = (1+x)
$$


That is, that the constant sequence converges trivially to $(1 + x)$