How to find the sum? Based on logarithm function expansion The problem:
How to find the sum?
$$-\sum_{i=1}^{\infty}\frac{(-x)^{i\; \bmod(k-1)}}{i}$$
Details:
I tried find this sum using Mathematica
-Sum[((-x)^(Mod[i,k-1]))/i,{i,1,Infinity}]
but I got no answer.
This sum is based on the $\ln(x+1)=-\displaystyle\sum_{i=1}^{\infty}\frac{(-x)^i}{i}$
The final result that I need is formula with finite indices, something like $-\displaystyle\sum_{j=0}^{k-2}f(x,k)$, so how to find this $f(x,k)$?
This another question shows exactly what I need.
Unfortunally I could't find how to multisect this function and find this sum until now.
 A: Details like mod $k-1$ or $-x$ are unnatural and pointless brain-pain so I think I'll ignore them.
$$\sum_{n=1}^{(m+1)k}\frac{u^{n\mod k}}{n}=\sum_{n=1}^k\left(\frac{1}{n}+\frac{1}{n+k}+\frac{1}{n+2k}+\cdots+\frac{1}{n+mk}\right)u^{n-1}$$
$$=\sum_{n=0}^{k-1}\left(\int_0^1 t^n+t^{n+k}+\cdots+t^{n+mk}dt\right)u^n=\int_0^1\frac{1-(ut)^k}{1-ut}\frac{1-t^{mk+1}}{1-t^k}dt.$$
I think the only way the limit $m\to\infty$ exists is if $u$ is a $k$-th root of unity so that there is a cancellation, otherwise the $(1-t^k)^{-1}$ factor will be an insurmountable singularity at $t=1$. Note the $u=0$ case is divergent when we understand $0^0$ to be $1$. If $u^k=1$ we then easily obtain
$$\int_0^1\frac{dt}{1-ut}=\frac{1}{u}\log\left(\frac{1}{1-u}\right).$$
I haven't thought about the justification for the $u^k=1$ restriction; don't have time at the moment.

Eh, note that if $u^k=1$ then the original power series is just that of $-u^{-1}\log(1-u)$ anyway (because $u^n=u^{n\mod k}$), so all the other stuff is unnecessary and the real point of concern should be proving the series only converges when $u^k=1$.
