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I have seen here that if we sign the terms in the harmonic series in a pattern of $p$ positive terms followed by $n$ negative terms and so on, without rearrangement, then the series converges iff $p=n$.

I was trying to figure out the limit in the case $p=n$. I got it for $p=n=2$ by subtracting the alternating odd harmonic terms, which converge to $\frac{\pi}{4}$, and the remaining terms were half the alternating harmonic series, so the result is $\frac{\ln{2}}{2}+\frac{\pi}{4}$. In a similar way (but using WolframAlpha) I did it for $p=n=3$, and the result was $\frac{\ln{2}}{3}+\frac{2\pi}{3\sqrt{3}}$. However, I don't know how to obtain a general result.

LRG1993
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  • Has it been proven that $p=n$ is necessary AND sufficient for the convergence ? – Peter Mar 06 '22 at 12:50
  • I thought so, from what I read for example here: https://math.stackexchange.com/questions/742643/change-of-signs-in-harmonic-series – LRG1993 Mar 06 '22 at 12:55

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I now got the right idea, so I want to give my insight, however I am not sure if this is the right way to do it in a comment.

By considering the power series for $\frac{x^k}{1+x^n}$ for every $0 \leq k \leq n-1$, and applying Abel’s theorem, we can show that $\sum_{m=0}^{\infty} \frac{(-1)^m}{nm+k}=\int_0^1 \frac{x^k}{1+x^n}dx$ (for $k=n-1$ we get $\sum_{m=0}^{\infty} \frac{(-1)^m}{m+1}=\int_0^1 \frac{x^{n-1}}{1+x^n}=\frac{\ln{2}}{n}$). With this in hand, we can write the desired series as the sum of $n$ such series, and so the total sum is $\int_0^1 \frac{1+x+x^2+...+x^{n-1}}{1+x^n}dx$. I don't know a way to simplify this expression, but I am satisfied with it.

LRG1993
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