The power series $\sum_{n=1}^\infty \ln(n)z^n$ has radius of convergence $1$ and $z=1$ is a singular point. Is $z=1$ an isolated singularity? If yes, what kind of isolated singularity?

I am only able to deduce that $z=1$ cannot be a pole.

Such type of questions appear naturally when one tries to relate the singularities of the power series and those of the Dirichlet series associated to the same sequence.


1 Answer 1


Let $f(z)$ be your function. Then $g(z)=f(z)(1-z)$ is equal to $$ (1-z)\sum_n \ln(n)z^n=\sum_{n\geq 2} (\ln(n)-\ln(n-1))z^n $$

Now, $\ln(n)-\ln(n-1)=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$, which gives us $$ g(z)=\sum_{n≥1} \left(\frac{1}{n}+g_n\right)z^n, $$

where $g_n=\ln(n)-\ln(n-1)-\frac{1}{n}=O\left(\frac{1}{n^2}\right)$ for $n>1$ and $g_1=-1$, so that $$ f(z)=\frac{\ln(1-z)}{z-1}+\frac{h(z)}{z-1}. $$

Here $h(z)$ is holomorphic in the unit disc and continuous on its boundary. Hope this answers your question on the type of singularity in $z=1$.


Let me also present a slightly less elementary way to study properties of this series, based on my favorite method of lots of contour integration. We will use the derivative of Riemann zeta-function, so this is more in the spirit of the question.

Let $x\in \mathbb C$ be a number with positive real part. Using the formula

$$ e^{-nx}=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} \Gamma(s)(nx)^{-s}ds, $$

we obtain

$$ f(e^{-x})=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty}\Gamma(s)\zeta'(s)x^{-s}ds. $$

From this we easily get

$$ f(e^{-x})=\mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}+\sum_{n\geq 0}\mathrm{Res}_{s=-n}\,\Gamma(s)\zeta'(s)x^{-s}. $$

The first summand is actually a bit different from all the other, because we get double pole. From expansions

$$ \Gamma(s)=1-\gamma(s-1)+O((s-1)^2), \zeta'(s)=\frac{1}{(s-1)^2}+O(1) $$


$$ x^{-s}=\frac{1}{x}-\frac{(s-1)\ln x}{x}+O((s-1)^2) $$

(here $\gamma$ is the Euler-Mascheroni constant) we get

$$ \mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}=-\frac{\ln x+\gamma}{x}, $$

which corresponds to the first part of my answer and also gives $h(1)=-\gamma$. The rest is way easier to compute and we obtain

$$ f(e^{-x})=-\frac{\ln x+\gamma}{x}+\sum_{n\geq 0}\frac{(-1)^n\zeta'(-n)x^n}{n!}. $$

Now, from this answer about derivative of zeta we see that this series has a nonzero radius of convergence (namely, $2\pi$) and we can even see singularities at $x=2\pi i n$ for $n\in \mathbb Z$, which is of course what one should expect because of singularity of $f$ at $z=1$.

  • 1
    $\begingroup$ Hello! So just to make sure, the singularity is a pole, right? $\endgroup$
    – Blue
    Jan 20, 2020 at 15:18
  • 1
    $\begingroup$ @Blue, you mean at $2\pi i n$? No, this is still a logarithmic singularity (would be super strange if $f(e^{-x})$ had a logarithmic singularity at $x=0$ but only a pole at $x=2\pi i$), because the factor $\psi(2n)$ in $\zeta'(1-2n)$ grows logarithmically in $n$ and for $x=2\pi i$ we get essentially the same series with logs $\endgroup$ Jan 20, 2020 at 17:03
  • 1
    $\begingroup$ I think that the second part of the answer is more satisfactory. The first part, while it reveals a branch point at $z=1$, does not clarify whether $h(z)$ is holomorphic in a neighborhood of $z=1$. The second explanation, if the statement about the radius of convergence checks out, does not have this flaw. $\endgroup$
    – Bogdan Ion
    Jan 21, 2020 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.