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Let $a_n$ be a sequence of real numbers satisfying $\sum_{n=1}^\infty |a_n-a_{n-1}|<\infty$. Then the series $\sum_{n=0}^\infty a_nx^n;x\in \Bbb R$ is convergent

  • nowhere in $\Bbb R$.
  • everywhere in $\Bbb R$.
  • on some set containing $(-1,1)$.
  • only on $(-1,1)$.

Since $\sum_{n=1}^\infty |a_n-a_{n-1}|<\infty\implies \lim|a_n-a_{n-1}|=0\implies \lim (a_n-a_{n-1})=0$

So if I take $a_n=\dfrac{1}{n}$ then the hypothesis holds.

Then corresponding to my chosen $a_n$ ; $\sum_{n=0}^\infty a_nx^n$ will converge if I choose $|x|<1$ since then we will have a series of the form $\sum \frac{1}{n}x^n$ where $|x|<1$ which will converge.

So I think option $4$ is correct. However, I can't prove it.

Please check if it's true and please give some hints on how to prove it.

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    $\begingroup$ There is the Cauchy-Hadamard's theorem $\endgroup$ – Michael Rozenberg Jun 22 '17 at 5:37
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The hint:

Use the Cauchy-Hadamard's theorem

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The answer is "there is not enough information to tell." In the case $a_n \equiv 1$ the series converges exactly on $(-1, 1)$; in the case $a_n = 1/n!$ it converges everywhere in $\mathbb{R}$.

Also, technically option 4 automatically implies option 3. Poorly phrased problem.

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  • $\begingroup$ You just gave examples that both satisfy option 3, thus showed that only option 3 is a valid answer. $\endgroup$ – Hagen von Eitzen Jun 22 '17 at 5:57
  • $\begingroup$ @HagenvonEitzen Fair point, but I think it's still an ambiguously phrased problem. It should say something like "which of these options is true for all sequences $a_n$ satisfying the hypothesis?". I think that with the current phrasing, you could reasonably argue to a teacher that you assumed $a_n \equiv 1$ and chose option 4. For a multiple-choice problem, all but one possibility should be unambiguously false. If nothing else, math classes should teach the importance of mathematical clarity. $\endgroup$ – tparker Jun 22 '17 at 6:16

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