# a GATE question on radius of convergence

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 $$r = \dfrac{1}{\lim\limits_{n\rightarrow \infty}|c_{n}|^\frac{1}{n}}.$$

• 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? – Actually Fritz Mar 27 at 1:17
• $a_{n}$ is $4^{(-1)^{n} n}$ – Unaccustomed Left-hander Mar 27 at 1:21
• 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? – Varun Vejalla Mar 27 at 1:56
• @varun it is either $4^{-1}$ or  4^{1}. Which one should I choose? – Unaccustomed Left-hander Mar 27 at 2:10
• 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, $|a_m|^{1/m}$. – 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, $$R_{\text{convergenge}}=\sqrt \frac 14=\frac 12$$