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 $i \in \mathbb{N}$?



Can a certain sum converge to 0?
[Yes](http://www.sciencedirect.com/science?_ob=MImg&_imagekey=B6WKD-4CRP4J8-DW-1&_cdi=6904&_user=145269&_pii=0022314X73900486&_origin=gateway&_coverDate=06/30/1973&_sk=999949996&view=c&wchp=dGLbVtz-zSkzV&md5=1b36190b46571879e3ff62c0490aa20e&ie=/sdarticle.pdf). 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.


