Find the radius of convergence of the power series $$\sum_{n=0}^{\infty} 4^{(-1)^{n} {n}}z^{2n}.$$

My attempt: Since the radius of convergence $r$ is given by $$r = \dfrac{1}{\lim\limits_{n\rightarrow \infty}|c_{n}|^\frac{1}{n}}.$$

Please help me further.

  • 1
    $\begingroup$ The radius of convergence of a power series $\sum_{n \geq 0} a_n z^n$ is defined as the reciprocal of $\limsup_{n \to \infty} \vert a_n \vert^{\frac{1}{n}}$ (I do not know what "lt" stands for). What is $a_n$ in your case? Can you try computing it? $\endgroup$ – Actually Fritz Mar 27 at 1:17
  • $\begingroup$ $a_{n}$ is $ 4^{(-1)^{n} n} $ $\endgroup$ – Unaccustomed Left-hander Mar 27 at 1:21
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    $\begingroup$ Using that $a_n = 4^{(-1)^nn}$, what does $|a_n|^{1/n}$ simplify to? Then what is the $\limsup\limits_{n \to \infty}$ of that expression? $\endgroup$ – Varun Vejalla Mar 27 at 1:56
  • $\begingroup$ @varun it is either $ 4^{-1}$ or $ $ 4^{1}. Which one should I choose? $\endgroup$ – Unaccustomed Left-hander Mar 27 at 2:10
  • $\begingroup$ It's the limit superior, which is $\lim_{n \to \infty} \left(\sup_{m \ge n}x_m\right)$, where $x_m$ is, in this case, $|a_m|^{1/m}$. $\endgroup$ – Varun Vejalla Mar 27 at 3:07

The given series converges if $\limsup |4^{(-1)^{n} {n}}z^{2n}|^\frac 1n=\limsup|4^{(-1)^{n}}||z^2|\lt 1$
$|a_n|^\frac 1n=1/4,4,1/4,\cdots$

Therefore by $\limsup$ definition, $\lim \sup |a_n|^\frac 1n=4$ so $|z^2|\lt \frac 14$ and therefore, $R_{\text{convergenge}}=\sqrt \frac 14=\frac 12$


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