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Prove that if we choose signs for individual terms in harmonic series $\sum_{n=1}^{\infty}{1\over n}$ in such a way that $p$ positive terms are followed by $q$ negative terms (without rearranging the series), we get divergent series for $p \neq q$ and convergent series for $p = q$.

I think that using $ \ln n = \sum_{n=1}^{\infty}{1 \over n} - \gamma - \epsilon_n$ I should get a series in terms of logarithm involving $p$ and $q$, but I just can't find it...

mirgee
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    Did you try to study the cases (p,q)=(1,1) and (p,q)=(2,1), to see what happens? – Did Apr 06 '14 at 22:14
  • @Did: Yes, I did. I know what happens, just don't know how to write it formally. – mirgee Apr 07 '14 at 06:26
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    If you "don't know how to write formally" a proof for the case (p,q)=(1,1), THAT case should be your question! – Did Apr 07 '14 at 06:36
  • @Did 2: I do, but I want general case, and then the indeces get wild! That's why I'm asking for help. – mirgee Apr 07 '14 at 09:22
  • Sorry but I fail to see your written proofs of the cases (p,q)=(1,1) and (p,q)=(2,1). Where are they? – Did Apr 07 '14 at 12:26
  • @Did: When p=q, the highest powers in the numerator cancel out, otherwise not, so by ratio test, the series converges/diverges. Is that what you are hinting at? – mirgee Apr 07 '14 at 15:03
  • No--in the sense that I do not know what you are referring to (what ratio test? which "highest powers"?). Let me suggest once again to fully solve the (p,q)=(1,1) and (p,q)=(2,1) cases before embarking on the general case. – Did Apr 07 '14 at 20:57
  • With two harmonic series, taking $p$ positive terms from one and $q$ negative terms from the other, the result is $\log\left(\frac{p}{q}\right)$. For example, $$\log\left(\frac{3}{2}\right)=1+\frac{1}{2} +\frac{1}{3}- 1 - \frac{1}{2} + \frac{1}{4}+\frac{1}{5}+\frac{1}{6} - \frac{1}{3} -\frac{1}{4}+...$$ – Jaume Oliver Lafont Oct 12 '21 at 05:52

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This is easily answered via summation by parts, e.g. using Abel's summation formula. If we let $\varepsilon_n$ be the sign we give to the term $\frac{1}{n}$, then note that

$$E(x) = \sum_{n \leqslant x} \varepsilon_n$$

remains bounded for $p = q$ - we have $0 \leqslant E(x) \leqslant p$ then - and for $p \neq q$ we have $E(x) \sim \frac{p-q}{p+q}\cdot x$ - if $k(p+q) < x \leqslant (k+1)(p+q)$, then $k(p-q) - q \leqslant E(x) \leqslant k(p-q) + p$. In both cases, we see that $b(x) := E(x) - \frac{p-q}{p+q}\cdot x$ is a bounded function on $[1,\infty)$.

Thus by Abel's formula we have

\begin{align} \sum_{n \leqslant x} \frac{\varepsilon_n}{n} &= \frac{E(x)}{x} + \int_1^x \frac{E(t)}{t^2}\,dt \\ &= \frac{p-q}{p+q} + \frac{b(x)}{x} + \frac{p-q}{p+q}\int_1^x \frac{dt}{t} + \int_1^x \frac{b(t)}{t^2}\,dt \\ &= \frac{p-q}{p+q}\cdot\log x + \Biggl(\frac{p-q}{p+q} + \int_1^{\infty} \frac{b(t)}{t^2}\,dt\Biggr) + \frac{b(x)}{x} - \int_x^{\infty} \frac{b(t)}{t^2}\,dt \\ &= \frac{p-q}{p+q}\cdot\log x + \Biggl(\frac{p-q}{p+q} + \int_1^{\infty} \frac{b(t)}{t^2}\,dt\Biggr) + O(1/x). \end{align}

Hence we have convergence for $p = q$ and divergence like $\frac{p-q}{p+q}\cdot \log x$ for $p\neq q$.

Daniel Fischer
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