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I have no idea to find all subintervals $[0,\infty)$ on which $\sum_{n=0}^{\infty}\left(\frac{x}{x-2}\right)^n$ converges uniformly or pointwise

Using ratio test, I can show the series of $f_n(x)$ is convergent in $0\le x <1$. How can I continue to show uniform or pointwise convergence

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  • $\begingroup$ Do you know the difference between pointwise and uniform? $\endgroup$ – Don Larynx Nov 22 '13 at 2:09
  • $\begingroup$ basically, yes. $\endgroup$ – Hugh C. Nov 22 '13 at 2:10
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Definition 1 Suppose $f_n$ is a sequence of functions sharing the same domain and codomain (for the moment, we defer specifying the nature of the values of these functions, but the reader may take them to be real numbers). The sequence $f_n$ converges pointwise to $f$, often written as

$$\lim_{n\rightarrow\infty}f_n=f\ \mbox{pointwise},$$

if and only if

$$\lim_{n\rightarrow\infty}f_n(x)=f(x).$$

Uniform convergence is a type of convergence stronger than pointwise convergence.

Definition 2 A sequence $g_n$ of functions converges uniformly to a limiting function $g$ if the speed of convergence of $g_n(x)$ to $g(x)$ does not depend on $x$.

Thus the easiest way to find out if it's uniform or not is to simply suppose you have a $y \neq x$. Then if $y$ does not change the limit $\lim_{n\to\infty}\{x_n\}$ of the sequence, you have a uniformly convergent sequence.

Hint if $x = 3$ then the series diverges. If $x = 0.1$ then it doesn't diverge (WHY?). Find the radius of (pointwise) convergence.

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  • $\begingroup$ Your definition of uniform convergence is vague. Can you not use the usual definition? $\endgroup$ – Pedro Tamaroff Nov 22 '13 at 2:22
  • $\begingroup$ Pedro, to be more rigorous, A sequence of functions $\{f_n\}, n = 1, 2, 3, \dots$ is said to be uniformly convergent to $f$ for a set $E$ of values of $x$ if, for each $\epsilon > 0$, an integer $N$ can be found such that $|f_n(x) - f(x)| < \epsilon$ for all $x \in E$ and $n \geq N$. $\endgroup$ – Don Larynx Nov 22 '13 at 2:25
  • $\begingroup$ @DonLarynx by using ratio test, I can get $\frac{|x|}{|x-2|}<1$ and then for $x>0$, the convergence radius must be $0\le x<1$, but how can i continue to show it is unif or pointwise? $\endgroup$ – Hugh C. Nov 22 '13 at 2:45
  • $\begingroup$ Take two different points in your radius $x$ and $y$ where $x$ is in the range of $f$ and $y$ in $g$'s range. Then they both have different limits. However the functions $f_n$ and $g_m$ both converge point wise to $x$ and $y$ respectively.. It is not uniform convergent then. $\endgroup$ – Don Larynx Nov 22 '13 at 2:57

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