# Singularities of power series

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.

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 $$z=1$$.

• Hello! So just to make sure, the singularity is a pole, right?
– Blue
Jan 20, 2020 at 15:18
• @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 Jan 20, 2020 at 17:03
• 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. Jan 21, 2020 at 17:42