When will be the series $\sum_{n=0}^\infty a_nx^n;x\in \Bbb R$ converge

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.

• There is the Cauchy-Hadamard's theorem – Michael Rozenberg Jun 22 '17 at 5:37

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 $\mathbb{R}$.
• @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. – tparker Jun 22 '17 at 6:16