endpoints convergence after integrating/mutiplying /subtracting power series 
*

*when we multiply a power series that converges for all values of $x$ by another power series of interval of convergence $(-1,1]$, then the new interval of convergence is the intersection of the 2 intervals which is $(-1,1]$? Do we have to check convergence at $x=\pm 1$? Sometimes after multiplication, we got a series that we can't put in closed form so we can't apply ratio test to check the endpoints?

*When we subtract/add  a power series with interval of convergence $(-1,1)$ from/to another power series of interval of convergence $(-1,1]$, the new interval of convergence is the intersection. Do we have to check endpoints?

*When we integrate or differentiate a power series, if the endpoints are included in the interval of convergence before integrating or differentiating the series, do they be included for the new power series?

*if we have a power series with interval of convergence $(-1, 1]$, and if we replace $x$ by $4x$ in the power series , then the new interval of convergence will be $(-0.25,0.25]$?
 A: *

*This involves a rearrangement of an iterated double summation into a single summation. Rearrangements and reductions of summation order are both fraught with dangers. Without careful examination, I would not even guarantee that the product converges on $(-1,1)$, much less at $x = \pm 1$. 

*As gt6989b has said, you can be sure that the sum diverges for $x = 1$. If it did not, then the divergent series at $1$ could be written as the difference of two convergent series, and thus would converge after all. At $x = -1$, however, all bets are off. It could be that the two divergent series sum to a divergent series, or the cause of the divergences could cancel and they would sum to a convergent series.

*For differentiation, this is definitely false. A simple counterexample is $$\sum_{n=0}^\infty \frac{x^n}{n^2}$$
This converges for $x = 1$, but the derivative obviously does not. Integration is much tamer in its effects. However, I believe it is still possible for the original series to converge at $x = 1$, but the integral does not. I don't have a fully worked out counterexample, but here is the idea: suppose that $k$ adjacent elements $a_i$ are positive and add up to some amount $\epsilon$, then the $k+1$st element is negative and brings the total back to $0$. Continue this pattern, except that the amount $\epsilon$ drops with each repetition, eventually converging to $0$. The sum of the $a_i$ converges to $0$ as well. But if you take the integral of the power series, dividing each $a_i$ by $i$, then that $k+1$ element is reduced in size more than the previous $k$ elements and now no longer cancels them out. By choosing carefully, it should be possible to make this residual large enough that the integrated series does not converge.

*yes. $4x \in (-1,1]$ if and only if $x \in (-1/4, 1/4]$.

