I want to find out the radius of the following power series of a complex variable:

$\displaystyle \sum_{-\infty}^{\infty} 3^{-|n|} z^{2n}, z \in \mathbb{C}$

The ration test $\displaystyle \lim_{n \to \infty}\left|\frac{a_m}{a_{m+1}}\right|$ gives me the radius of convergence as 3 . But I am not really sure how to handle the negative power terms. Should I group same co-efficient terms, and complete the square? But then, I don't get a polynomial in a single variable.

Also, how do I evaluate the sum?


1 Answer 1


Your series is $$ 1 + \sum_{n=1}^\infty \bigg(\frac{z^2}3\bigg)^n + \sum_{n=1}^\infty \bigg(\frac1{3z^2}\bigg)^n, $$ which converges when both $|z^2/3|$ and $|1/3z^2|$ are less than $1$, that is, when $1/\sqrt3<|z|<\sqrt3$.

This is typical of Laurent series, which are convergent on annuli and thus have both inner and outer "radii of convergence".

  • $\begingroup$ One might not want to encourage the OP to use the term "radius of convergence" to an object to which it does not apply. $\endgroup$
    – Did
    Mar 31, 2013 at 8:32
  • $\begingroup$ Woudn't this be just a necessary condition for convergence? How can I prove that it actually converges, for e.g. by finding the sum? $\endgroup$
    – user23238
    Mar 31, 2013 at 8:54
  • $\begingroup$ @Did I don't find it damaging to say that a Laurent series, convergent on an annulus, has an inner and outer radius of convergence. At most it's just a matter of terminology. $\endgroup$ Mar 31, 2013 at 19:29
  • $\begingroup$ @ramanujan_dirac Both sums are geometric series, and we know everything there is to know about when geometric series converge. Indeed, you can find the sum exactly in this case; but that's not necessary in general for knowing that a series converges. $\endgroup$ Mar 31, 2013 at 19:30

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