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Let $\{r_i\}_{i \in \aleph}$ be sequence of integers such that, for some $t \in \mathbb{N}$ and all $i \in \mathbb{N}$, we have $r_i = r_{i+t}$. My question:

Can $\displaystyle \sum_{i=1}^n \dfrac{r_i}{i}$ converge to $0$ if $n \rightarrow \infty$ for a non-trivial choice of the $r_i$ and $t$? Or does $\displaystyle \sum_{i=1}^\infty \dfrac{r_i}{i} = 0$ imply $r_i = 0$ for all $i \in \mathbb{N}$?

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Stylistic hint: unless you really mean Aleph, the LaTeX code for natural numbers is \mathbb{N}. –  darij grinberg Apr 4 '11 at 23:46
    
Also, minor remark: your result easily yields that $\frac{\pi}{\ln 2}$ is irrational, so we should not expect anything TOO easy to work. –  darij grinberg Apr 4 '11 at 23:47
    
(By "your result" I mean "$r_i=0$ for all $i$".) –  darij grinberg Apr 4 '11 at 23:48
    
The sum can be expressed as a linear combination of values of $L(1,\chi)$, the Dirichlet $L$-function, over various characters $\chi$, with coefficients coming from a cyclotomic extension of the rationals. There are results about the non-vanishing of $L(1,\chi)$; maybe there are results about the non-vanishing of linear combinations. –  Gerry Myerson Apr 4 '11 at 23:59
    
And if I'm not mistaken, it also implies the non-zeroness of the Dirichlet L-series for $s = 1$, which is not TOO easy either –  Woett Apr 5 '11 at 0:01
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2 Answers 2

up vote 22 down vote accepted

Yes. All the $r_i$ must equal $0$ if the period is prime, however. Consider for example $$f(s)=(1-p^{1-s})^2 \zeta(s),$$ which is periodic with period $p^2$, at $s=1$.

I should probably expand on this answer a bit. The case where $t$ is prime is an old conjecture of Chowla, which was resolved by Baker, Birch, and Wirsing (all the $r_i=0$ in this case) in the paper I link to in the first word of this answer. They give the Dirichlet series for $f(s)$ above as a counterexample when $t$ is not prime.

To see that $f(s)$ has the desired properties, I'll work it out in a bit more detail for $p=2$. Expanding $f$ out as a Dirichlet series gives $$f(s)=\sum_{n=0}^\infty \frac{1}{(4n+1)^s}-\frac{3}{(4n+2)^s}+\frac{1}{(4n+3)^s}+\frac{1}{(4n+4)^s}$$ as Woett remarks in the comments. On the other hand, $(1-2^{1-s})^2$ has a double zero at $s=1$, whereas the zeta function $\zeta(s)$ has a simple pole at $s=1$; so $f(1)=0$. So taking the limit as $s\to 1^+$ gives that the OP's series converges to $f(1)=0$ for $r_1=r_3=r_4=1,~ r_2=-3, ~t=4$, as desired.

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So in particular, $r_1 = r_3 = r_4 = 1, r_2 = -3, t = 4$ answers my question. Neat! –  Woett Apr 5 '11 at 0:43
    
Yup! Baker's method is pretty cool also; the paper (and Baker's theorem) is worth going through. –  Daniel Litt Apr 5 '11 at 0:48
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Another way to do the sum is to let $$g(x)=x-(3/2)x^2+(1/3)x^3+(1/4)x^4+(1/5)x^5-(3/6)x^6+\dots$$ and use the usual methods to show that $$g(x)=\log{x+1\over x^2+1}$$ which gives $g(1)=0$. –  Gerry Myerson Apr 5 '11 at 7:25
    
Edited to fix the broken link. –  Daniel Litt Apr 9 '11 at 22:33
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If $r_i=r_{i+t}$ for some $t$ and all $i$ you can split the sum into a finite sum and the rest $$\sum_{i=1}^t \frac{r_i}{i}+\sum_{i=t}^\infty \frac{r_i}{i}=\sum_{i=1}^t \frac{r_i}{i}+r_t\sum_{i=t}^\infty \frac{1}{i} $$ by your assumption. So the latter is just the harmonic series and your series diverges unless $r_t=0$.

This reduces your question to a finite question. Now the answer depends on what you call non-trivial. Eventually all $r_i$ have to be 0, but but of course solutions exist. For example $r_i=0$ for all $i \neq 1,2$ and set $r_1=1$ and $r_2=-2$.

EDIT: Ok, I misread the question. And my answer is not correct. At least using the arguments above you can show that the series can only be conditionally convergent. Since otherwise, you could rearrange the series as follows

$$ r_1 \sum_{i \equiv 1 \mod t}\frac{1}{i}+\cdots+r_{t-1} \sum_{i \equiv t-1 \mod t}\frac{1}{i}$$. And these are essentially the harmonic series.

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The condition is not that the $r_i$ are eventually constant; it is that they are periodic. –  Daniel Litt Apr 4 '11 at 23:40
    
wood, I don't think you are right. For example let $r_{2k}=0$, $r_{2k+1}=\left(-1\right)^k$, then the sum is $-\ln 2$. –  darij grinberg Apr 4 '11 at 23:45
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What we CAN say is that $r_1+r_2+...+r_t=0$. –  darij grinberg Apr 4 '11 at 23:45
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@all. yes you are completely right. I just made a quick edit to clarify things. I should have refrained from answering at all at 2 o'clock in the morning ;-) –  wood Apr 4 '11 at 23:53
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