By Dani Hobbes

2012-03-10 22:15:00 8 Comments

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.


@Pookaros 2017-05-10 23:18:42

Even though the 1st answer is the more easy and appropriate here is a more analytic one. $\sum_{n=1}^{k} (z^n)/n =(z/1-z) $$\sum_{n=1}^{k-1} [(1-z^n)/n(n+1) + (1-z^k)/k]$

(we can prove that with induction till k).

We can also see that

0 $\leqslant$ |(1-z^k)/k|$\leqslant$ 2/k$\rightarrow$0 for k$\rightarrow$$\infty$ also

0 $\leqslant$ |(1-z^n)/n(n+1)|$\leqslant$ 2/n(n+1)

$\sum_{n=1}^{\infty}2/n(n+1)$ $(converges)$ so $\sum_{n=1}^{\infty} (z^n)/n$ $(converges)$ as well

@Martin Argerami 2012-03-11 00:56:15

Fix $z$ in the unit circle, i.e. $|z|=1$. 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 $a_n=1/n$, $b_n=z^n$. The first two conditions are clearly satisfied, and for the third one: $$ \left|\sum_{n=1}^Nz^{ n }\right|=\left|\frac{z-z^{N+1}}{1-z}\right|\leq\frac{2}{|1-z|} $$ 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 $|z|>1$.

@philmcole 2017-12-06 12:32:40

Can you explain this step: $|\sum_{n=1}^Nz^{ n }| = |\frac{z-z^{N+1}}{1-z}|$ ?

@Martin Argerami 2017-12-06 13:33:37


@philmcole 2017-12-06 13:48:17

Yeah, but where do you get that additional $z$ in the nominator from? By the formula you linked it should be $|\frac{1-z^{N+1}}{1-z}|$, no?

@Martin Argerami 2017-12-06 13:49:24

If you start the sum from $0$, yes.

@Pacciu 2012-03-11 23:23:01

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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