Uniformly convergence of power series I know when r is radius of convergence of power series
Power series is uniformly converge on Any closed interval of (-r , r)
I want know when power series converges on (-r , r] , [-r,r] or [-r,r) each case can include each endpoint of interval as a uniformly convergence domain?
 A: In these notes, Abel's theorem is given in the form
Theorem $6.5.4$: If $\sum_{n=0}^\infty a_nx^n$ converges at $x=R>0$, then the power series converges uniformly on $[0,R]$.  A similar result holds when the series converges at $x=-R<0$. 
The proof is given in an appendix, and is too long to include here.  I'm not familiar with Abel's theorem in this form.  I know it as asserting that if the series converges at an endpoint, then the limit function is continuous as $x$ approaches the endpoint from inside the interval of convergence.  I felt that Abel's theorem, in the form I know it, was the key to the proof, but I couldn't do it.  I'll have to read this proof.   
Anyway, the answer to your question is "yes."  
A: It is likely that you need to test those functions individually, as when deriving the formula for convergence radius, we used Cauchy's test, which is
$$\limsup _{n \to \infty} a_n ^{\frac 1 n} = c$$
1) If $c \lt 1$ the series converges
2) If $c \gt 1$ the series diverges
3) If $c = 1$ the test is inconclusive
Therefore there are examples where a power series converges absolutely or conditionally at both end point, conditionally at one and diverges at one, or diverges at both.
However if a power series is convergent on a end point, it is converges uniformly there.
Proof (WARNING: see comments) Suppose that this is not true. Without losing generality, suppose that a power series is convergent on $x_0+r$, but is not uniformly convergent.
Since it is uniformly convergent on every compact subset of $(x_0-r,x_0+r)$, this means that for any $\epsilon \gt 0$, the series is uniformly convergent on $[x_0,x_0+r-\epsilon]$.
Therefore it is not uniformly convergent on $[x_0+r-\epsilon,x_0+r]$, which means for some $\epsilon^* \gt 0$, there is no such a number $N \gt 0$ such that as long as $n \gt N$, for all $x \in [x_0+r-\epsilon,x_0+r]$, $\lvert F_n(x) - F(x) \rvert \lt \epsilon^* $.
Therefore for all number $N \gt 0$, there is some number $x^* \in [x_0+r-\epsilon,x_0+r]$, such that $\lvert F_n(x^*) - F(x^*) \rvert \gt \epsilon^*$.
Now let $N = 1, 2, 3, ...$ and we will get a list of $x^*_1, x^*_2, x^*_3, ...$. The collection $\{(x^*_i,x^*_{i+1})\}$ is a open covering of $[x_0+r-\epsilon,x_0+r]$, since the latter is compact, by applying Heine-Borel theorem, we can conclude that the power series is not convergent at $x_0+r$, which is a contradiction. $\square$
A: In order to find whether or not the power series is uniformly convergent on $[-r,r)\\$ or $(-r,r]\\$ or $[-r,r]\\$ you need to find that if the series is convergent at the end points i.e. $\sum a_n(-r)^n\\$ or $\sum a_nr^n\\$ or both are convergent respectively. 
This way you can include an end point to the interval of uniform convergence.
I hope it helps : )
