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 		 	Let $R$ denote   the radius of convergence   of the power series   $\sum\limits_{k=1}^\infty kx^k$.   Then

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 			   \item $R>0$ and     the series is convergent on    $[-R,R]$
 			  	\item $R>0$ and    the series converges at    $x=-R$    but does not converge at    $x=R$
 			  	\item $R>0$ and    the series does not converge  outside   $(-R,R)$.
 			  	\item $R=0$
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 	  {\bf{Answer}} \\
 	   We have    the radius of convergence    $R=\displaystyle \lim_{k\to \infty}\frac{a_k}{a_{k+1}}   =\lim_{k\to \infty}\frac{k}{{k+1}}   =\lim_{k\to \infty}\frac{1}{{1+\frac{1}{k}}}   =1.$\\
 	   Hence \textit{(4)} is wrong.\\
 	   For $x=R=1$,    $\displaystyle \sum k x^k=\sum k$, is    divergent. \\    Hence \textit{(1)} is wrong.\\
 	   For $x=-R=-1$,   \\ $\displaystyle \sum k x^k=\sum k (-1)^k   =-1+2-3+4\cdots   =(-1+2)+(-3+4)+(-5+6)+\cdots   =-1-1-1-\cdots$
 	\\
 	   Hence $\displaystyle \sum k (-1)^k$    is divergent.
 	   \\Therefore \textit{(2)} is wrong.\\
 	  Also,    since $R=1   >0$,    by the property of radius of convergence,    the series does not converge    outside $(-R,R)$.\\
 	  Hence \textit{(3)} is correct.
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