In what set of complex numbers does these series $(a_n)_{n=1}^{\infty}$ convergence/divergence? If series convergences is the convergence steady (uniform?) within the convergence area.

$(a)$ $a_n=\frac{(-1)^n z}{n}$

\begin{align} r&=\lim_{n\to\infty}\Bigg| \frac{\frac{(-1)^{n+1}}{n+1}}{\frac{(-1)^n}{n}}\Bigg| \\ &=\lim_{n\to\infty}\Bigg| \frac{(-1)^n (-1)^1}{n+1} \frac{n}{(-1)^n}\Bigg| \\ &=\lim_{n\to\infty}\Bigg|\frac{-n}{n+1}\Bigg|=|-1|=1 \end{align}

So it converges for $\forall z \in \mathbb{C}$ when $|z|<1$. Doesn't this mean it diverges if $|z|>1$?

$(b)$ $a_n=nz^n$

\begin{align} r=&\lim_{n\to\infty}\Bigg| \frac{(n+1)z^{n+1}}{nz^n} \Bigg| \\ &=\lim_{n\to\infty}\Bigg| \frac{(n+1)z^nz}{nz^n} \Bigg| \\ &=\lim_{n\to\infty}\Bigg| \frac{nz+z}{n} \Bigg| \\ &=\lim_{n\to\infty}\Bigg| z+\frac zn \Bigg| \\ &=|z| \end{align} Converges when $|z|<1$ for all $z\in\mathbb{C}$.

$(c)$ $a_n=\frac{z^n}{n}$

\begin{align} r&=\lim_{n\to\infty}\frac{\sqrt[n]{|z^n|}}{\sqrt[n]{n}} \\ &=\frac{|z|}{1}=|z| \end{align} Converges when $|z|<1$ for all $z\in\mathbb{C}$.

  • $\begingroup$ I think you are misunderstanding the question. It is about series $\sum_{n\ge 1}a_n$ and not about the convergence of the sequence $\lim_{n\to \infty}a_n.$ $\endgroup$ – mfl Oct 5 '14 at 17:20
  • $\begingroup$ Hmmm. What approach should I use then? $\endgroup$ – ELEC Oct 5 '14 at 17:27
  • 1
    $\begingroup$ You may find useful the convergence tests in this link en.wikipedia.org/wiki/Convergent_series $\endgroup$ – mfl Oct 5 '14 at 17:34

$$\lim_{n\to\infty }\left|\frac{a_{n+1}}{a_n}\right|=1$$ where $a_n=\frac{(-1)^n}{n}$. Then the serie converge for all $z\in\mathbb C$ such that $|z|<1$.

  • $\begingroup$ I edited my OG post. Can I use this ratio test on the others as well? $\endgroup$ – ELEC Oct 5 '14 at 17:52
  • $\begingroup$ what is OG ? Of course, you can ;-) $\endgroup$ – idm Oct 5 '14 at 18:02
  • $\begingroup$ original post. I tried what you did there to $a_n=nz^n$ and it didn't work out so well. Root test gave me $r=z$. $\endgroup$ – ELEC Oct 5 '14 at 18:10
  • $\begingroup$ (b) and (c) came to the same result $r=z$ $\endgroup$ – ELEC Oct 5 '14 at 18:18
  • $\begingroup$ It doesn't !! $a_n=n$ and not $nz^n$. But you can also take $a_n=nz^n$, then $$\lim_{n\to\infty }\left|\frac{a_{n+1}}{a_n}\right|=|z|$$ and you know by d'Alembert theorem that $$\sum_{n=1}^\infty a_n$$ converge if and only if $$\lim_{n\to\infty }\left|\frac{a_{n+1}}{a_n}\right|=\rho<1$$ who gives precisely in your exemple that $|z|<1$. $\endgroup$ – idm Oct 5 '14 at 18:18

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