# Finding the Radius of Convergence of Series.

Find the radius of convergence of the series: $$\sum_{n=0}^{\infty} 5^{(-1)^n}z^n$$

This was an old homework question. I tried finding the radius of convergence by using the ratio test, but the professor wrote on my homework that the ratio test does not work. I, therefore, am not sure how to or what method to use to find the radius of convergence. The answer should be $1$ according to the back of the book. Any help, suggestions, and pointers are welcomed. I have attached the question as a picture. Thank you.

You can split the summation in

$$5(1+z^2+z^4+z^6+\cdots)+\frac15(z+z^3+z^5+z^7+\cdots).$$ This is allowed as all terms are positive.

The radius of convergence of the sum will be the smallest of the two radii.

When it converges, the value of the sum is

$$\frac5{1-z^2}+\frac z{5(1-z^2)}.$$

• What are the two radii specifically? – kemb Apr 2 '17 at 20:03
• @bob: your job to find that. – Yves Daoust Apr 2 '17 at 20:04
• Isn't one of the radii 5 and the other 1/5? – kemb Apr 2 '17 at 20:05
• @BOB: no, use the theory. (And if that helps, consider $t=z^2$.) – Yves Daoust Apr 2 '17 at 20:06
• I'm confused, could you please explain, what theory? – kemb Apr 2 '17 at 20:06

For $|z|=1$, the sequence $(5^{(-1)^n}z^n)$ is neither convergent to $0$ nor unbounded, so radius of convergence is $1$.

Another way to look at it : if $|z|<1$, $\left|5^{(-1)^n}z^n\right|\le 5|z|^n$ which is the term of a convergent geometric series, so radius is greater than $1$, but if $|z|>1$, $\left|5^{(-1)^n}z^n\right|\le \frac15|z|^n$ which is the term of a divergent geometric series.