Convergence radius of $\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ \left\lfloor \sqrt { n } \right\rfloor } }{ n } } { x }^{ n }$

Question

I need to find the radius of the seria

$$\sum _{ n=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ \left\lfloor \sqrt { n } \right\rfloor } }{ n } } { x }^{ n }$$

where, $\left\lfloor \sqrt { n} \right\rfloor $ is floor function.I haven't got any idea.Any help will be appriciated

Answer 1

Use Cauchy root test,$$\frac { 1 }{ R } =\overline { \underset { n\rightarrow \infty }{ lim } } \sqrt [ n ]{ \left| \frac { { \left( -1 \right) }^{ \left\lfloor \sqrt { n } \right\rfloor } }{ n } \right| } =\lim _{ n\rightarrow \infty }{ \frac { 1 }{ \sqrt [ n ]{ n } } =1 } $$,so $$\left| x \right| <1$$