Is it known that $\sum_{i=1}^\infty \frac{if\;(i \pmod n=0)\;then\;(1-n)\;else\;(1)}{i}=log(n)$?

Question

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.

Answer 1

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

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