0
$\begingroup$

I found an unclear step for me in Zorich, Mathematical Analysis I, sec. 5.5, pag. 270. We are trying to find all $z \in\mathbb{C}$ for which the series:

$$c_0+c_1(z-z_0)+c_2(z-z_0)^2+...$$

converges. To do this, we try to understand when it converges absolutely, applying the Cauchy criterion to the series:

$$|c_0|+|c_1(z-z_0)|+|c_2(z-z_0)^2|+...$$

obtaining that it converges if:

$$|z-z_0|< \frac{1}{\overline{\lim}_{n\to\infty}\sqrt[n]{|c_n|}} $$

and all is ok. Now, Zorich says (referring to the series with absolute values): the general term does not tend to zero if $|z-z_0| \geq \frac{1}{\overline{\lim}_{n\to\infty}\sqrt[n]{|c_n|}} $.

I can't completely understand this sentence, in particular the reason why he also includes the case $|z-z_0| =\frac{1}{\overline{\lim}_{n\to\infty}\sqrt[n]{|c_n|}} $. In this case I would have:

$$|c_0|+\frac{|c_1|}{\left (\overline{\lim}_{n\to\infty}\sqrt[n]{|c_n|} \right )} +\frac{|c_2|}{\left( \overline{\lim}_{n\to\infty}\sqrt[n]{|c_n|}\right )^2} +...$$

and so I would verify that:

$$\lim_{k\to\infty}\frac{|c_k|}{\left( \overline{\lim}_{n\to\infty}\sqrt[n]{|c_n|}\right )^k}\neq 0$$

but I can't. How does Zorich manage to say that for all $z\in\mathbb{C}$, such that $|z-z_0| =\frac{1}{\overline{\lim}_{n\to\infty}\sqrt[n]{|c_n|}} $, the series with absolute values diverges?

Thanks.

$\endgroup$
  • 1
    $\begingroup$ If you've faithfully reproduced the sentence, then he is wrong. For the series $\sum_{n=1}^\infty \frac 1nz^n$, the limit is $1$, but the general term for $|z| = 1$ does converge to $0$. And the series itself is well-known to diverge when $z = 1$ and converge when $z = -1$. $\endgroup$ – Paul Sinclair Feb 24 at 20:11
  • $\begingroup$ Thank you for your feedback. $\endgroup$ – Nameless Feb 24 at 21:39

Your Answer

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

Browse other questions tagged or ask your own question.