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  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 $(-0.25,0.25]$?

gt6989b
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MCS
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  • i think for 1,2, the endpoints don't have to be checked; since it diverges for the first series, the result will diverge also unless it's a special power series, e.g. the zero function :). (4) is correct, (3) I am not sure. – gt6989b Mar 15 '16 at 20:57

1 Answers1

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  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 $x \in (-1/4, 1/4]$.
Paul Sinclair
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  • I was curious about the scenario you mentioned in 1 (product fails to converge on $(-1,1)$). This answer suggests that this failure cannot occur. Unfortunately, the proof was given in a link to notes by Pete Clark, and the link is broken. –  Mar 16 '16 at 00:33
  • Nice example where the radius of convergence of the product is larger than the radii of the factors: http://math.stackexchange.com/a/500166/169852 –  Mar 16 '16 at 00:37
  • @Bungo - I assume that is correct, even without seeing the proof. I was only saying that on general principles, I would not count on it. My own limited investigation indicated it would work if the ratio test was applicable to all three, but that of course cannot be assumed. – Paul Sinclair Mar 16 '16 at 01:48
  • If $a_n x^n$ converges for some value of $x$, then certainly $\frac{a_n}{n+1}x^{n+1}=x\left(\frac{a_n x^n}{n+1} \right)$ would converge. I've seen in many textbooks "by direct comparison" cited incorrectly (the $a_n$ need not be positive), but certainly by Dirichlet test. Regardless, simply looking at the form of a term-by-term differentiated or integrated series shows that differentiation can not gain endpoints, and integration can not lose endpoints. – Marcel Besixdouze Oct 19 '18 at 19:12