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

• the final series does not converge – qbert Sep 15 '16 at 18:38
• "For which $x \ge 0$ does the power series ... converge uniformly?" Sorry but this is most unclear since it makes no sense to say that a series converges uniformly at some point $x$. Are you asking to find a set on which the series converges uniformly? – Did Sep 15 '16 at 19:21

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.

• oh, it was a geometric series the whole time. That was tricky for me to see. – user369299 Sep 15 '16 at 19:39
• By the way, at $x=0$, the series converges but not uniformly, am I correct? – user369299 Sep 15 '16 at 19:45
• it happens. But in general, what helps me is bounding the terms above by series that are easy, which you could do here – qbert Sep 15 '16 at 19:45
• at $x=0$ it's just zero right? – qbert Sep 15 '16 at 19:46
• Yes, hence why I asked. I know, that was a very trivial thing to ask... – user369299 Sep 15 '16 at 19:48

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

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

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