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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

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    $\begingroup$ The convergence radius is unaffected by the funny power of $-1$. The endpoints could be, but that does not seem to be asked about. $\endgroup$ – André Nicolas Jul 27 '16 at 17:01
  • $\begingroup$ That is not a power series because the coefficients depend on $x.$ You can still talk about the values for which this series converges, but "radius of convergence" is not really the right term here. Also, what does $\sqrt x$ mean if $x$ is negative? $\endgroup$ – zhw. Jul 27 '16 at 17:03
  • $\begingroup$ is there $\\ \\ \sqrt { n } $ or $\\ \\ \sqrt { x } $?,becuase i posted aswer as $\\ \\ \sqrt { n } $ $\endgroup$ – haqnatural Jul 27 '16 at 17:14
  • $\begingroup$ It was typo, sorry $sqrt(n)$ $\endgroup$ – user356640 Jul 27 '16 at 17:19
  • $\begingroup$ @AndréNicolas you could say that it is unaffected by the whole $\frac{(-1)^{\lfloor \sqrt{n} \rfloor}}{n}$ term $\endgroup$ – reuns Jul 27 '16 at 18:33
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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 $$\left| x \right| <1$$

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