Uniform convergence of a series

Question

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 $$ \sum_{n=0}^\infty \frac{(n-1)^{n-1}}{n^n} < \infty, $$ so that, using the Weierstrass M-test, the power series converges uniformly for all $x \ge 0$?

Answer 1

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 $$ \frac{1}{1+x}<1\Rightarrow 1<1+x\Rightarrow x>0 $$ 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.

Answer 2

$$ \sum_{n=0}^\infty \frac x{(1+x)^n} = x\sum_{n=0}^\infty \frac {1}{(1+x)^n}= x\sum_{n=0}^\infty (1+x)^{-n}= x(\frac{1}{x}+1)= (1+x) $$

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

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