The series I have is $$\displaystyle\sum_{n=0}^\infty {\dfrac{z^n}{1+z^{2n}}}$$

The same series with absolute values is: $$\displaystyle\sum_{n=0}^\infty {\dfrac{|z|^n}{1+|z|^{2n}}}$$

Using D'Alembert's principle, $$\displaystyle\lim {\dfrac{a_{n+1}}{a_n}} = {\dfrac{|z|^n \cdot |z|}{1+|z|^{2n} \cdot |z|}} \cdot {\dfrac{1+|z|^{2n}}{|z|^n}} = |z|$$

The convergence range is when $|z| < 1$. But the book answer is $|z| \ne 1$.

  • $\begingroup$ What's $Z_n$? Is it actually $Z^n$? $\endgroup$ – xbh Nov 18 '18 at 9:33
  • $\begingroup$ Yes, sorry. Updated $\endgroup$ – user3132457 Nov 18 '18 at 9:34
  • $\begingroup$ I don't think the calculation of $\lim a_{n+1}/a_n$ is correct. $\endgroup$ – xbh Nov 18 '18 at 9:36
  • $\begingroup$ So what is wrong? $\endgroup$ – user3132457 Nov 18 '18 at 9:38
  • $\begingroup$ Wouldn't $|z|>1$ and $|z|<1$ lead to different results? $\endgroup$ – xbh Nov 18 '18 at 9:41

If $|z|=r<1$, and $n\ge 1$, then $$ \left|\frac{z^n}{1+z^{2n}}\right|\le \frac{|z|^n}{1-|z|^{2n}}=\frac{r^n}{1-r^{2n}} <\frac{r^n}{1-r} $$ and hence the series $$ \sum_{n=0}^\infty\frac{z^n}{1+z^{2n}} $$ converges, due to Comparison Test.

If $|z|=1$, and in particular $z=i$, then the series is not even definable.

Note. This is not a power series, and hence finding the radius of convergence is out of question. Clearly, there exist values of $z$, with $|z|>1$, for which the series converges absolutely, i.e., all $z\in\mathbb R$, with $|z|>1$. Meanwhile, the unit circle is a natural boundary of the series, since, for the points $z=\exp(ik/2^\ell)$ are singularities (not isolated) of the series, for all $k,\ell\in\mathbb N$.

  • $\begingroup$ I don't get how $$\left|\frac{z^n}{1+z^{2n}}\right|\le \frac{|z|^n}{1-|z|^{2n}}$$? If $z^{2n}$ is greater than 1, then the inequality doesn't hold. $\endgroup$ – user3132457 Nov 18 '18 at 9:56
  • $\begingroup$ This is true when $|z|=r<1$, as mentioned in the beginning of the answer. $\endgroup$ – Yiorgos S. Smyrlis Nov 18 '18 at 10:00
  • $\begingroup$ I know, but the answer is $|z| \ne 1$ which means $|z| > 1$ is possible to have. $\endgroup$ – user3132457 Nov 18 '18 at 10:02

$$a_n\left(\frac1z\right)=\frac{1/z^n}{1+1/z^{2n}}\\ =\frac{z^n}{z^{2n}+1}\\ =a_n(z)$$ So it converges for $z$ whenever it converges for $1/z$.


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