Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|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
1  
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

1 Answer 1

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

$$\int_0^1\frac{dt}{1-ut}=\frac{1}{u}\log\left(\frac{1}{1-u}\right).$$

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 $u^k=1$.

share|improve this answer
    
(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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.