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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 $x \ge 0$?

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  • $\begingroup$ the final series does not converge $\endgroup$ – qbert Sep 15 '16 at 18:38
  • $\begingroup$ "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? $\endgroup$ – Did Sep 15 '16 at 19:21
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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.

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  • $\begingroup$ oh, it was a geometric series the whole time. That was tricky for me to see. $\endgroup$ – user369299 Sep 15 '16 at 19:39
  • $\begingroup$ By the way, at $x=0$, the series converges but not uniformly, am I correct? $\endgroup$ – user369299 Sep 15 '16 at 19:45
  • $\begingroup$ it happens. But in general, what helps me is bounding the terms above by series that are easy, which you could do here $\endgroup$ – qbert Sep 15 '16 at 19:45
  • $\begingroup$ at $x=0$ it's just zero right? $\endgroup$ – qbert Sep 15 '16 at 19:46
  • $\begingroup$ Yes, hence why I asked. I know, that was a very trivial thing to ask... $\endgroup$ – user369299 Sep 15 '16 at 19:48
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$$ \sum_{n=0}^\infty \frac x{(1+x)^n} = x\sum_{n=0}^\infty \frac {1}{(1+x)^n}= x\sum_{n=0}^\infty (1+x)^{-n}= x(\frac{1}{x}+1)= (1+x) $$

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

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

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