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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): enter image description here 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)

enter image description here

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.

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    $\begingroup$ This cannot exist, by summation by parts: if $A_n=\sum_{k\le n} a_k$, then $\sum_{n\le N} a_n/n$ is essentially $A_N/N+\sum_{n\le N} A_n/n^2$, which converges. $\endgroup$ Dec 4, 2021 at 0:16
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    $\begingroup$ More precisely, $\sum_{n=1}^N a_n/n$ equals $A_N/N+\sum_{n=1}^{N-1} A_n/(n^2+n)$, which converges. $\endgroup$
    – GH from MO
    Dec 4, 2021 at 0:56
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    $\begingroup$ A fun question... but/and, as it happens, the disappointing answer is that no such crazy thing can exist. "Too bad!" :) $\endgroup$ Dec 4, 2021 at 2:32
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    $\begingroup$ @ChristianRemling The summation by parts is a really nice approach. It also makes it clear that even if we bound that size of $a_k$ (for instance, insisting $|a_k| < 1$), we can still make the sum $\sum \frac{a_k}{k}$ arbitrarily large (just make $A_n$ large for a long time by having the negative terms appear arbitrarily late in the sequence), but never infinite since $A_n$ has to go to zero eventually. Since we've only shifted terms a finite distance away, we can still keep $\sum a_k = 0$ as well. Very cool! $\endgroup$ Dec 4, 2021 at 2:57
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    $\begingroup$ Christian's argument is known as Dirichlet convergence test en.m.wikipedia.org/wiki/Dirichlet%27s_test $\endgroup$ Dec 4, 2021 at 8:11

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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.

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  • $\begingroup$ Hmm, I might be interpreting the equations wrong, but I'm not sure if the first equal sign is true. And the final result that $| \sum \frac{a_n}{n} | \leq |A(x)|$ seems false, since defining $a_0 = 1$, $a_1 = -1$ and 0 for the rest of $a_n$ gives $A(x) = 0, x \geq 2$, but $1 + (-1)/2 = 1/2$ $\endgroup$ Dec 4, 2021 at 3:11
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    $\begingroup$ (there should be a max! specifically when bounding the integral you pulled out A(x) and not the maximum of A on [1,x] —- of course it’s all ok because A is bounded!) $\endgroup$
    – alpoge
    Dec 4, 2021 at 3:33
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    $\begingroup$ @alpoge Ah your right, I was just thinking that’s what was missing. Then the max pops out of the integral and it makes sense again, thank you! $\endgroup$ Dec 4, 2021 at 3:38
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    $\begingroup$ @CalebBriggs: Well, the max thing is also clear from my remark (which is Christian Remling's remark made precise). As $\sum_{n=1}^N a_n/n$ equals $A_N/N+\sum_{n=1}^{N-1} A_n/(n^2+n)$, its absolute value is at most $\max_{n\leq N}|A_n|$. $\endgroup$
    – GH from MO
    Dec 4, 2021 at 5:09
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    $\begingroup$ In any case $\sum a_n$ has bounded partial sums, so this is exactly an application of Dirichlet test as observed by Fedor Petrov in comments (and there is no need to repeat the proof) $\endgroup$ Dec 4, 2021 at 10:08

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