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The problem:

How to find the sum?

$$-\sum_{i=1}^{\infty}\frac{(-x)^{i\; \bmod(k-1)}}{i}$$


I tried find this sum using Mathematica


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.

share|cite|improve this question
Why do you (a) always use $i$ as an infinite sum index and (b) mod things out $k-1$ instead of just $k$? You're giving me OCD or something. – anon Nov 18 '11 at 23:30
@anon ^^ is because to my final formula I need to use the $k-1$ index , but I agree if $l=k-1$ and $k=l+1$ is the same problem. About $i$ to infinity, well, just happens =) – GarouDan Nov 18 '11 at 23:34
If $x\ne x^2$, by doing a little bit of arithmetic with fractions you can break the sum into blocks of $k-1$ terms that are all positive (or all negative). Convergence would imply absolute convergence, whence you can rearrange the sum. It's easy to do so to obtain a logarithmically divergent series, contradicting the assumption of convergence. That's why Mathematica was reticent: the sum doesn't converge for real $x$ except when $x=0$ or $x=1$. – whuber Nov 19 '11 at 0:02
@whuber: How should we define $x^0$ here when $x=0$? – anon Nov 19 '11 at 0:23
@anon That's a good point: if we define $0^0=1$, the series diverges. But we can make perfect sense of the sum by adopting a slightly different set of representatives of the equivalence classes modulo $k-1$ than you might be thinking of: instead of $0, 1, \ldots, k-2$, use $1, 2, \ldots, k-1$. :-) – whuber Nov 19 '11 at 0:58
up vote 3 down vote accepted

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


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 $u^k=1$.

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(Also, this formula generates the well-known example $\log2=1-1/2+1/3-1/4+1/5-\cdots$.) – anon Nov 19 '11 at 0:06
Thx again @anon ^^. Great solution. – GarouDan Nov 30 '11 at 1:08

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