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Inspired by the exponential series, I'm curious about where exactly the series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges.

I calculated $$ \limsup_{n\to\infty}\sqrt[n]{\frac{1}{n}}=\limsup_{n\to\infty}\frac{1}{n^{1/n}} $$ and $$ \lim_{n\to\infty} n^{1/n}=e^{\lim_{n\to\infty}\log(n)/n}=e^{\lim_{n\to\infty}1/n}=e^0=1. $$ So the radius of convergence is $1$, so the series converges on all $z$ inside $S^1$. But is there a way to tell for which $z$ on the unit circle the series converges? I know it converges for $z=-1$, but diverges for $z=1$, but I don't know about the rest of the circle. For what other $z$ does this series converge? Thanks.

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From the Wikipedia articles on the Mercator series (en.wikipedia.org/wiki/Mercator_series) and Abel's Test (en.wikipedia.org/wiki/…), it converges everywhere on the circle but $z=1$. –  Steven Stadnicki Mar 10 '12 at 22:22
    
Thanks @StevenStadnicki. I see how Abel's test applies here now, but I don't see how the article on the Mercator series is relevant. What exactly are you using in that article? –  Hobbie Mar 10 '12 at 22:29
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I think this would be a good opportunity to take a look at summation by parts. –  Sam Mar 10 '12 at 22:41
    
@Hobbie Specifically, the bit under 'Complex series' where it describes where the series converges and how Abel's test gets you there. :-) Note that since the series they gave there is $f(z) = \ln(1+z)$, your series is $-f(-z) = -\ln(1-z) = \ln(\frac{1}{1-z})$. –  Steven Stadnicki Mar 10 '12 at 22:43

2 Answers 2

up vote 8 down vote accepted

For any $z$ in the unit circle, $z=e^{2\pi i t}$ with $t\in[0,1]$. We want to apply Dirichlet's test: if $\{a_n\}$ are real numbers and $\{b_n\}$ complex numbers such that:

  1. $a_1\geq a_2\geq\cdots$
  2. $\lim_{n\to\infty}a_n=0$
  3. There exists $M>0$ such that $\left|\sum_{n=1}^Nb_n\right|\leq M$ for every $N\in\mathbb{N}$;

then $\sum_{n=1}^\infty a_nb_n$ converges. Here $a_n=1/n$, $b_n=z^n=e^{2\pi i n t}$. The first two conditions are clearly satisfied, and for the third one: $$ \left|\sum_{n=1}^Ne^{2\pi i n t}\right|=\left|\frac{e^{2\pi i t}-e^{2\pi i (N+1)t}}{1-e^{2\pi i t}}\right|\leq\frac{2}{|1-e^{2\pi i t}|} $$ for all $N\in\mathbb{N}$. This shows that the third condition is satisfied for every $z\ne1$ in the circle.

In conclusion, the series converges for every $z$ with $|z|\leq1$ other than $z=1$, and it diverges for $|z|>1$.

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The followin theorem on power series is due to E. Picard:

Let $(a_n)$ be a sequence of real numbers.

If the sequence $(a_n)$ is nonnegative, decreases and tends to zero when $n\to \infty$, then the complex power series $\sum a_n\ z^n$ converges in the closed unit disc $\overline{D}(0;1)$ with the only possible exception of the point $1$.

The proof of Picard's theorem relies on Abel's summation by parts formula, as far as I remember.

Now, the coefficients of your series, i.e. $a_n=1/n$, satisfy the assumptions of Picard's theorem, hence your series converges at least in $\overline{D}(0;1)\setminus \{1\}$; on the other hand, the series diverges when $z=1$ (for it becomes the harmonic series).

Therefore the convergence set of $\sum 1/n\ z^n$ is $\overline{D}(0;1)\setminus \{1\}$.

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