Answer:
(-1,1).
Step-by-step explanation:
We need to calculate [tex]\lim_{n \to \infty}\frac{| a_{n+1}|}{| a_{n}|} = \frac{1}{R}[/tex] where R is the radius of convergence.
[tex]\lim_{n \to \infty}\frac{\frac{|(-1)^{n+1}|}{|n+1|}}{\frac{|(-1)^{n}|}{|n|}}[/tex]
[tex]\lim_{n \to \infty}\frac{\frac{1}{|n+1|}}{\frac{1}{|n|}}[/tex]
[tex]\lim_{n \to \infty}\frac{|n|}{|n+1|}[/tex]
Applying LHopital rule we obtaing that the limit is 1. So [tex]1=\frac{1}{R}[/tex] then R = 1.
As the serie is the form [tex](x+0)^{n}[/tex] we center the interval in 0. So the interval is (0-1,0+1) = (-1,1). We don't include the extrem values -1 and 1 because in those values the serie diverges.