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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.

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1 Answer 1

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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 $z=1$.

Edit:

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

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

and

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

(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 $z=1$.

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    $\begingroup$ Hello! So just to make sure, the singularity is a pole, right? $\endgroup$
    – Blue
    Jan 20, 2020 at 15:18
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    $\begingroup$ @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
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    $\begingroup$ 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

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