1
$\begingroup$

The sum $$\sum\limits_{k=1}^{\infty}\frac{z^k}k$$ is given and I want to find out for which $z\in\mathbb C$ the series converges. By applying the ratio test or calculating the radius of convergence, we learn that the series converges for $|z|<1$ and diverges for $|z|>1$.

But what about the case $|z|=1$? If $z=1$, for example, the harmonic series does not converge, however, for $z=-1$ the series converges (alternating series test), so the convergence behaviour isn't the same for all $|z|=1$.

I've already tried writing $z$ as $a+\rm i\sqrt{1-a^2}$, but could not get anything helpful from that. How should I proceed?

$\endgroup$
3
  • 1
    $\begingroup$ The series converges for all $z\ne 1$. $\endgroup$
    – Mark Viola
    Commented Dec 1, 2016 at 16:35
  • 2
    $\begingroup$ You'll want to use the Dirichlet test here $\endgroup$
    – Ben Grossmann
    Commented Dec 1, 2016 at 16:36
  • $\begingroup$ WLOG $z=e^{2ix}$ $$\dfrac{z^k}k=\dfrac{(e^{2ix})^k}k$$ $$S=\sum_{k=1}^\infty\dfrac{z^k}k=-\ln(1-e^{2ix})=-\ln\{-e^{ix}(e^{ix}-e^{-ix})\}$$ For $\pi>x>0,$ $$S=-\ln(-i)-ix+\ln(2\sin x)=-i\pi-\dfrac{i\pi}2-ix+\ln(2\sin x)$$ $\endgroup$
    – lab bhattacharjee
    Commented Dec 1, 2016 at 16:40

1 Answer 1

Reset to default
2
$\begingroup$

Hint. Write $z = e^{it}$, where $t$ is not an integral multiple of $2\pi$ (so that $z \neq 1$). So we consider the series $\sum\limits_{k = 1}^\infty \frac{e^{ikt}}{k}$. Show that the partial sums $\sum\limits_{k = 1}^N e^{ikt}$ are bounded by $\lvert \csc(t/2)\rvert$. Conclude that the series converges by Dirichlet's test.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .