Can a certain sum converge to 0? 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 $i \in \mathbb{N}$?
 A: 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.
A: 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.
