I'm hoping to find an explicit construction for a sequence such that $\sum a_n = 0$ but $\sum \frac{a_n}{n} = \infty$, or a proof that one cannot exist. So far, I have a good idea of how we can increase the value of the sum up to infinity. For example, we could construct a sequence like this (a_n shown in red):
It stays constant for a while, and then that constant part is canceled by a large negative term. Because the flat part becomes smaller and smaller, the partial sums (shown in green) will eventually go to zero.
We now have that $\sum \frac{a_n}{n} > 0$, since the negative terms occur later, they will be smaller than the positive terms that occurred earlier. However, this method can only get something finite, more is needed to get the sum all the way to infinity.
Another thought is that we could rearrange the sum, but flipping the order in which the negative terms appear. For instance, here I reverse the order of every group of 5 negative terms, so that larger negative terms occur later (flipped shown in black)
I'm fairly certain this is still insufficent to get out an infinite value (mainly since the effect will be neglible on the tail, so the partial sums of $\sum \frac{a_n}{n}$ still converge). What's interesting, is that this rearrangement hasn't effected the value of the original sum, we still have $\sum a_n = 0$.
Now, I know that by the Reimann rearrangement theorem, there is some rearrangement $\sigma (n)$ such that $\sum a_n = 0$ but $\sum \frac{a_{\sigma(n)}}{\sigma(n)} = \infty$ if $\frac{a_n}{n}$ is conditionally convergent. But, what I'm truly curious about, is if there is some $a_n$ such that $\sum a_{n} = 0$ and $\sum \frac{a_{n}}{n} = \infty$ when we are forced to evaulate the sum without rearranging (so, we sum the n=1 term, then the n=2 term, then the n=3 term, etc.).
I'm thinking this is possible, perhaps by reversing a larger and larger amount of consecutive terms, up to infinity at the tail, and being careful about making sure the original sum still converges, though I don't have a good idea of one would actually do this correctly.
Best Answer
This is just a more detailed version of Christian Remling's comment. Define, for each positive integer $n$, the partial sum $$\displaystyle A(n) = \sum_{k \leq n} a_k.$$ By the definition of limits of series, it follows that $0 = \lim_{n \rightarrow \infty} A(n)$. By abuse of notation, we also denote by $A(x)$ to be $\sum_{k \leq x} a_k$ even when $x$ is not a positive integer. Then partial summation gives $$\displaystyle \left \lvert \sum_{n \leq x} \frac{a_n}{n} \right \rvert = \left \lvert \frac{1}{x} A(x) + \int_1^x \frac{1}{t^2} A(t) dt \right \rvert \leq \frac{|A(x)|}{x} + |A(x)| \left(1 - \frac{1}{x} \right) = |A(x)|,$$ which tends to $0$ as $x \rightarrow \infty$. Actually, we just need that the right hand side is bounded. This suffices for the proof.