Suppose we have an alternating series of the form $$\sum_{n=1}^\infty \frac{(-1)^{a_n}}{n}$$ We know this converges for some basic cases, like when $a_n=n$ (the sum evaluating to $-\ln2$) or for interesting alternating patterns like $$\sum_{n=1}^\infty \frac{(-1)^{\lfloor\frac{n}{2}\rfloor}}{n}=1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots=\frac{\pi}{4}-\frac{\ln2}{2}$$ I also recall seeing a question on here about an alternation resembling a factorial pattern, something like $$1-\underbrace{\left(\frac{1}{2}+\frac{1}{3}\right)}_{2!\text{ terms}}+\underbrace{\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\right)}_{3!\text{ terms}}-\cdots$$ which may or may not have converged, I cannot seem to find it...

Is there a generalization that can be made about the conditions on $a_n$ for these types of series?

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    $\begingroup$ I suspect a general resolution of this is a Hard Problem™. $\endgroup$
    – Chappers
    Sep 25, 2015 at 21:33
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    $\begingroup$ @JoseArnaldoBebitaDris No. The limit above is $1$ for any real $a_n$. $\endgroup$
    – Winther
    Sep 25, 2015 at 22:16
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    $\begingroup$ If the density of the set $\{n\colon a_n \text{ is even} \}$ exists, it must equal $\frac12$ for the series to converge. (But there are lots of sequences where that density doesn't even exist....) Even the density equaling $\frac12$ isn't sufficient, though: take $a_n=n$ for all $n$ except those of the form $2\lfloor k\log k\rfloor$ for some $k$, in which case take $a_n=-1$. Here the density equals $\frac12$ but the series diverges to $-\infty$. $\endgroup$ Sep 25, 2015 at 22:23

1 Answer 1


I don't think you're going to find a characterization that's not just a disguised version of "the series converges if and only if the series converges"; convergence of this series is a ding an sich (sp?), seems to me.

But the series does converge for "most" choices! If $(\epsilon_n)$ is a sequence of plus or minus ones chosen "at random" then the series $\sum \epsilon_n/n$ converges almost surely.

(I was about to give a more precise statement of that, decided that would be silly; anyone with the background to understand the precise statement should be able to formulate it for himself. Regarding proving it, look in some probability book in the neighborhood of the Law of Large Numbers...)

  • $\begingroup$ I realize I made the "mistake" of over-generalizing. I have a specific set of cases I'm thinking about for $a_n$, but I'd like to think about it some more before I post another question on the topic. $\endgroup$
    – user170231
    Oct 10, 2015 at 0:58

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