[Math] endpoints convergence after integrating/mutiplying /subtracting power series

power series

  1. 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?

  2. 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?

  3. 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?

  4. 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 $(-0.25,0.25]$?

Best Answer

  1. 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 $x = \pm 1$.
  2. 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.
  3. 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.
  4. yes. $4x \in (-1,1]$ if and only if $x \in (-1/4, 1/4]$.
Related Question