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What is radius of convergence of series: $$\sum_{m=0}^{ \infty} \frac{(-1)^m}{8^m}(x^{3m})$$

I know that for a holomorphic function $f$ whose power series has coefficient $a_n$ is given as

$$\frac{1}{R}= \lim_{x \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$

Using this I am getting $8$, if I am not wrong. Please help.

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    Call $x^3=z$. Your series becomes $\sum_m (1/8^m)z^m$, whose radius of convergence is $8$. This means that $|x^3| \le 8$, .i.e. $|x| \le 2$. – 2017-01-26

1 Answers 1

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We have $a_n= \frac{(-1)^m}{8^m}(x^{3m})$.

Now $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\leq1$

$\left|\frac{(-1)^{m+1}x^{3m+3}}{8^{m+1}}\times\frac{8^m}{(-1)^mx^{3m}}\right|\leq1$

$\left|\frac{(-1)x^3}{8}\right|\leq1$

$\left|x^3\right|\leq8$ i.e $\left|x\right|\leq2$

So,radius of convergence is $2$.