The Series multisection
$$\sum_{n=0}^{\infty}a_{mn+b} z^{mn+b}=\frac{1}{m}\sum_{k=0}^{m-1}\omega^{-kb}f(\omega^{k}z)$$
for $f(z)=\sum_{n=0}^{\infty}a_{n} z^{n}$ and $\omega=e^{\frac{2\pi i}{m}}$ is well known. However, I don't see many people writing down restrictions on the values of $m$ and $b$
For some background context, I was trying to apply it on $$\sum_{k=1}^{\infty}\frac{z^{n+k}}{n+k}$$ for natural numbers $n$. I would get the result as $-\log(1-z)$ which is obviously false.
After I derive the formula again I finally found the mistake
Consider $f(z)$ not $\sum_{n=0}^{\infty}a_{n} z^{n}$ but, $$f(z)=\sum_{n\in\mathbb{Z}}^{\infty}a_{n} z^{n}$$ where $a_n=0$ for all negative $n$
$\mathbf{Formula\, 1}\sum_{j=0}^{m-1}\omega^{jn}=\frac{1-\omega^{nm}}{1-\omega^n}=0$ if $n/m$ is not an integer
$=m$ if $n/m$ is an integer.
Note that the condition of it is $n/m$ an integer. Not positive integer
With this formula considered. Calculate
$$\sum_{k=0}^{m-1}\omega^{-kb}f(\omega^{k}z)$$ $$=\sum_{k=0}^{m-1}\omega^{-kb}\sum_{j\in\mathbb{Z}}a_j\omega^{kj}z^j=\sum_{j\in\mathbb{Z}}a_jz^j\sum_{k=0}^{m-1}\omega^{k(j-b)}$$
Applying $\mathbf{Formula 1}$ on the sum inside gives
$$=m\sum_{j\in\mathbb{Z}}a_j z^j,\,\, \mathrm{for} \,\, m|(j-b)\,\,\,\mathrm{and}\,\,\ 0 \,\, \mathrm{otherwise}$$
$$\sum_{k=0}^{m-1}\omega^{-kb}f(\omega^{k}z)=m\sum_{m|(j-b)}a_j z^j$$ $m|j-b\Rightarrow mn=j-b$ for any $n\in\mathbb{Z}$ Gives $$\sum_{n\in\mathbb{Z}}a_{mn+b}z^{mn+b}=\frac{1}{m}\sum_{k=0}^{m-1}\omega^{-kb}f(\omega^k z)$$
I think this explains why I got $-\log(1-z)$. The formula at the top is assuming $b$ is smaller than $m$
Is my understanding correct?