0
$\begingroup$

I'm trying to find region of convegence of the Laurent series $$ \sum_{n=-\infty}^\infty \frac{z^n}{2^{-n}+4^n}$$

I expand it to $$ \sum_{n=0}^\infty \frac{z^n}{2^{-n}+4^n} + \sum_{n=1}^{\infty} \frac{z^{-n}}{2^{n}+4^{-n}}$$

After I use ratio test to get ROC for the first term (analogously for the second but the problem is in limits) $$ \lim_{n\to\infty} \bigg(\frac{z^{n+1}}{2^{-n-1}+4^{n+1}}\times \frac{2^{-n}+4^{n}}{z^{n}}\bigg) = \lim_{n\to\infty} \frac{2^{-n}+4^{n}}{2^{-n-1}+4^{n+1}}z $$ and the problem for me here is to find the limit. If I understand how the limit is done for the first term, I will apply it for the second

$\endgroup$
  • $\begingroup$ will it be equal to 1/4? $\endgroup$ – Lautern Feb 25 '18 at 12:20
0
$\begingroup$

$$\lim_{n\to\infty} \frac{2^{-n}+4^{n}}{2^{-n-1}+4^{n+1}}|z| =\lim_{n\to\infty} \frac{2^{-3n}+1}{2^{-3n-1}+4}|z| =\frac{|z|}4$$ etc. So you need $|z|<4$ for convergence. You need to look at the negative $n$ part of the series for the inner radius of the annulus.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.