Which one of the following are true?[CSIR-June 2017] Let $\{a_n\}$ be a sequence of real numbers satisfying $\sum _{n=1}^{\infty }\left|a_n-a_{n-1}\right|<\infty $, then the series $\sum_{n=1}^\infty a_nx^n,x\in \mathbb R$ is convergent.
(a)nowhere on $\mathbb R$
(b)Everywhere on $\mathbb R$
(c) On some set containing $(-1,1)$
(d)Only on $(-1,1)$
My Try
(a) False, $a_n=1/n$, $\sum_{n=1}^\infty x^n/n,x\in \mathbb (-1,1)$ is convergent
(b) False,  $a_n=1/n$, $\sum_{n=1}^\infty x^n/n,x\in \mathbb (-1,1)$ is convergent. Not converges everywhere.
(c) I think (c) is also false, $a_n=1/2^n$ $\sum_{n=1}^\infty x^n/2^n,x\in \mathbb (-1/2,1/2)$. which is not an interval containing (-1,1)
(d)False, $a_n=2^n$ $\sum_{n=1}^\infty 2^nx^n,x\in \mathbb (-2,2)$
 A: Your argument for (d) is not valid because $a_n=2^{n}$ does not satisfy the hypothesis. Note that $|a_n-a_m| \leq |a_n-a_{n+1}|+...+|a_{m-1}-a_m|$ for $n >m$. Conclude that $(a_n)$ Ia Cauchy sequence, hence bounded. It follows that the series $\sum a_n x^{n}$ converges in $(-1,1)$. Being a  power series it converges in some interval containing $(-1,1)$. Other parts of your answer are correct. 
A: You can solve this problem using the root criterion as follows.
First note that the sequence (a_n) is convergent:


*

*$a_n = a_1 + \sum_{k=1}^{n-1}(a_{k+1}-a_k)\stackrel{\sum_{k=1}^{\infty}|a_{k+1}-a_k|<\infty}{\Longrightarrow} \exists \lim_{n\to \infty }|a_n| =L\geq 0$
Now, there are two cases for the radius of convergence $\rho$ of $\sum_{k=1}^{\infty}a_nx^n$:
$$L>0 \stackrel{\forall n>N_{\epsilon:\;}0<\sqrt[n]{L-\epsilon} < \sqrt[n]{|a_n|} < \sqrt[n]{L+\epsilon}}{\Rightarrow} \lim_{n\to\infty}\sqrt[n]{|a_n|}=1\Rightarrow \rho = 1$$
$$L=0 \stackrel{\forall n>N_{\epsilon:\;}0\leq \sqrt[n]{|a_n|} < 1}{\Rightarrow} \limsup_{n\to\infty}\sqrt[n]{|a_n|}\leq 1\Rightarrow \rho \geq 1$$
So, the radius of convergence is at least $1$, hence the series converges on some set containing $(-1,1)$.
