find the radius of convergence of the power series $\sum _1^{\infty} n^{-\sqrt n} z^n $

here $a_n= n^{-\sqrt n}\\$

$ \frac{1}{R}=\lim_{n \rightarrow \infty} [ n^{-\sqrt n} ]^\frac{1}{n}$ for further i didnt get any one can help


Hint: $$(n^{-\sqrt{n}})^{1/n}=n^{-1/\sqrt{n}}=e^{\frac{-\log n}{\sqrt{n}}}$$

  • $\begingroup$ @boon het..sorry i didt get your idea can you please explain $\endgroup$ – gggg Sep 7 '16 at 8:19
  • $\begingroup$ All you have to do is compute the limit on the right hand side. I gave you a hint on how to approach it. $\endgroup$ – b00n heT Sep 7 '16 at 8:20
  • $\begingroup$ yes but i dont understand how u wirte in exponential function $\endgroup$ – gggg Sep 7 '16 at 8:21
  • $\begingroup$ $a=e^{\log a}$ the apply the property $\log(a^b)=b\log a$ $\endgroup$ – b00n heT Sep 7 '16 at 8:21

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