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I found this general infinite sum:

$\sum_{i=1}^\infty \frac{ \mathtt i \mathtt f \; (i \pmod n = 0) \; \mathtt t \mathtt h \mathtt e \mathtt n \;(1-n) \; \mathtt e \mathtt l \mathtt s \mathtt e \; (1)}{i} = log(n)$

Sample with n = 2:

log(2) = 1 - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + $\frac{1}{5}$ - $\frac{1}{6}$ + $\frac{1}{7}$ - ...

(the minus are given at (i MOD 2)=0 --> (1 minus 2) = -1)

Sample with n = 3:

log(3) = 1 + $\frac{1}{2}$ - $\frac{2}{3}$ + $\frac{1}{4}$ + $\frac{1}{5}$ - $\frac{2}{6}$ + $\frac{1}{7}$ + $\frac{1}{8}$ - $\frac{2}{9}$ + ...

(the minus are given at (i MOD 3)=0 --> (1 minus 3) = -2)

and so on.

Is this formula already known? Can that be proven?

Thanks in advance.

Chatton
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    Split the sum in two: 1) those $i$ where $i\equiv 0\mod n$ ($i=n,2n,3n,\ldots$) and 2) the rest (you have to the consider partial sums as one of these sums is divergent) – Winther Sep 16 '18 at 20:38
  • @Winther: surely if you split a convergent series as the sum of two series, the two series are either both convergent or both divergent.The latter happens in this case and I don't see how your approach can help. – Rob Arthan Sep 16 '18 at 20:52
  • @RobArthan Both series diverge. This approach works: it amounts to showing that the sum of the first $N$ terms is $\sum_{i=1}^{N}\frac{1}{i} - \sum_{i=1}^{\lfloor N/n\rfloor}\frac{1}{i}$ which has a finite limit. I added the comment about considering partial sums as both these series diverge if considered individually – Winther Sep 16 '18 at 20:54
  • Yes this is known. For a (rather strong) hint, see https://en.wikipedia.org/wiki/Harmonic_number#Calculation – Did Sep 16 '18 at 21:20

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You have to start with partial sums. Choose some $M$ to get:

$$ S_M = \sum_{i=1}^{nM} \frac{if\;(i \pmod n=0)\;then\;(1-n)\;else\;(1)}{i}= \sum_{i=1}^{nM} \frac1i - n \sum_{i=1}^M \frac1{n i} = \sum_{i=1}^{nM} \frac1i - \sum_{i=1}^M \frac1{i} $$

Now observe the well-known limit (see here) $\lim_{k \to \infty}\sum_{i=1}^k \frac1{i} = \ln(k) + \gamma$, where $\gamma$ is the Euler–Mascheroni constant.

This gives

$$ S = \lim_{M \to \infty} S_M = \ln(nM) + \gamma - (\ln(M) + \gamma) = \ln(n) $$

Andreas
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