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It is well known that $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}... =\log(2).$$
If we consider the array: $T(n,k) = -(n-1)\; \text{ if }\; n|k, \;\text{ else } \;1,$

Starting:

$$\displaystyle T = \left( \begin{array}{ccccccc} +0&+0&+0&+0&+0&+0&+0&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&+1&+1&-3&+1&+1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&+1&+1&+1&+1&-5&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$


Is it true that $\displaystyle \log(n)=\sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}$$\;$?

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  • $\begingroup$ And why do you expect this to hold (e.g. you computed the partial sums up to a big number and they are close to the $\log$s...)? $\endgroup$
    – Marek
    Jun 19, 2011 at 23:36
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    $\begingroup$ Why's this tagged as number-theory? $\endgroup$ Jun 20, 2011 at 1:46
  • $\begingroup$ @Marek & @George Lowther: Perhaps not an answer to your questions, but this recurrence here: list.seqfan.eu/pipermail/seqfan/2011-June/014999.html , led me to the values of the Mangoldt function here: list.seqfan.eu/pipermail/seqfan/2011-June/015006.html , which in turn led me to the series above. $\endgroup$ Jun 20, 2011 at 8:07
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    $\begingroup$ We can see the formula for log(k) in page 136 by Lehmer (1975) matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf $\endgroup$ Feb 4, 2016 at 11:34
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    $\begingroup$ The periodicity of T(n,k) suggests Fourier analysis, and the result is remarkable: since it is a zero-mean delta train, it has all components but the constant. $$\log(n) = \sum_{k=1}^\infty \frac{1}{k} \sum_{j=1}^{n-1} e^\frac{2 \pi ijk}{n} $$ $\endgroup$ Feb 5, 2023 at 12:19

6 Answers 6

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You can write $T(n,k)=1-n1_{\{n\mid k\}}$. Then, for $\vert x\vert < 1$ look at the power series $$ \begin{align} \sum_{k=1}^\infty\frac{T(n,k)}{k}x^k&=\sum_{k=1}^\infty\frac{x^k}{k}-\sum_{k=1}^\infty1_{\{n\mid k\}}\frac{nx^k}{k}\\ &=\sum_{k=1}^\infty\frac{x^k}{k}-\sum_{k=1}^\infty\frac{x^{nk}}{k}\\ &=-\log(1-x)+\log(1-x^n)\\ &=\log\left(\frac{1-x^n}{1-x}\right)\\ &=\log(1+x+\cdots+x^{n-1}). \end{align}. $$ So, letting $x$ increase to 1, $$ \lim_{x\uparrow1}\sum_{k=1}^\infty\frac{T(n,k)}{k}x^k=\log n. $$ The fact that you can commute this limit with the summation to get $\sum_{k=1}^\infty T(n,k)/k$ follows from the fact the series converges uniformly (over $0 < x < 1$). You can show this by grouping together the consecutive positive terms where $n\nmid k$ to get a sequence with alternating signs and decreasing in magnitude. Then, truncating the series gives an error which is bounded by the following term. That is, $$ \left\vert\sum_{k=1}^{jn-1}\frac{T(n,k)}{k}x^k-\log(1+x+\cdots+x^{n-1})\right\vert \le \frac{-T(n,jn)}{jn}x^{jn}\le \frac1j. $$ Commuting the limit with a finite sum is no problem, so you get $$ \left\vert\sum_{k=1}^{jn-1}\frac{T(n,k)}{k}-\log n\right\vert\le\frac1j. $$

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    $\begingroup$ Very nice derivation from basic principles. $\endgroup$ Jun 20, 2011 at 1:43
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Yes. You can get the sums by differentiating the digamma function repeatedly. There is a good deal of information about the resulting polygamma functions, including series expressions, here. Your matrix version is a lot more visually arresting than the usual Dirac delta function formulation!

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  • $\begingroup$ I'd still like to know how one comes up with such series and a conjecture of what they should converge to. Any clues? $\endgroup$
    – Marek
    Jun 20, 2011 at 1:07
  • $\begingroup$ @Marek: (This is only speculation.) There is a long history of evaluation of $\Gamma$, $\Gamma'/\Gamma$, and their derivatives at special points. So the answers ($\log n$) may have come before the questions (series). $\endgroup$ Jun 20, 2011 at 1:51
  • $\begingroup$ @AndréNicolas When you say resulting polygamma functions, what do you mean? I am trying to find a relationship between n/LambertW(n)-1 and Stirling numbers of the second kind. In the derivative of the explicit formula for Stirling numbers of the second kind I get the polygamma function, according to Mathematica. How do you find the polygamma function in relation to this question about logarithms? $\endgroup$ Oct 18, 2013 at 12:09
  • $\begingroup$ After some experimenting, I find that $$\sum _{n=0}^{\infty } \left(\frac{x^{2 n+1}}{(2 n+1)^s}-\frac{x^{2 n+2}}{(2 n+2)^s}\right) = 2^{-s} \left(x \Phi \left(x^2,s,\frac{1}{2}\right)-\text{Li}_s\left(x^2\right)\right)$$ where $\text{Li}_s\left(x^2\right)$ is the PolyLog function and $\Phi \left(x^2,s,\frac{1}{2}\right)$ is the LerchPhi function. Is this what you meant in your answer above? $\endgroup$ Oct 18, 2013 at 12:16
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The formula seem to be extendable to fractional arguments of the log. The key is to rewrite the formula for $\log(x)$ as difference of two sums but to a common limit. So we can write $$ \begin{eqnarray} \log(x) &=& \lim_{n\to \infty} \sum_{k=1}^n {1 \over k} - x\sum_{k=1}^{\lfloor n/x \rfloor} {1 \over x k} \\ &=& \lim_{n\to \infty} \sum_{k=1}^n {1 \over k} - \sum_{k=1}^{\lfloor n/x \rfloor} {1 \over k}\\ &=& \lim_{n\to \infty} \sum_{k=\lfloor n/x \rfloor+1}^n {1 \over k} \end{eqnarray} $$ It seems to be a possible improvement to take the mean of the two sums when the initial index is either $ \lfloor n/x \rfloor $ or $ \lfloor n/x \rfloor +1 $ . So the final best (but not too complicated) approximation might be $$ \begin{eqnarray} w_n &=& \lfloor n/x \rfloor\\ \log(x) &=& \lim_{n\to \infty} {1\over2w_n} + \sum_{k=w_n+1}^n {1 \over k} \end{eqnarray} $$ However, for reasonable digits of precision one needs many many terms, so this might be only of formal interest.
Moreover, maybe the formula in this notation is also known; I vaguely think I've seen series-formulae involving the floor-function in this or related ways...

[update] There is one more...
To think of fractional summation-bounds suggests to consider integration instead of sums. So I tried $$ \log(x) = \lim_{n \to \infty} \int_{n/x} ^n \frac 1t dt $$ and then even $$ \log(x) = \lim_{n \to \infty} \int_n^{nx} \frac 1t dt $$ and after that even could let n finite...
and the perfect result (even for small n) $$ \log(x) \underset{n \gt 0}{=} \int_n^{nx} \frac 1t dt $$ suggests to look into wikipedia to see, who had noticed that first... ;-) and it's nice to see the identity of the integral-definition and the simple reformulation and generalization of your surprising patterns.

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With a slight change in notation, define the partial sum $s_n:=\sum_{k=1}^n\frac{T(a,k)}k$. Consider the 'blocked' subsequence $(s_{an}, n=1,2,\ldots)$ such that $s_{an}$ is the sum of the first $n$ blocks of $a$ terms: $$s_{an}=\sum_{k=1}^{an}\frac{T(a,k)}k=\sum_{k=1}^{an}\frac1k-\sum_{j=1}^n\frac a{aj}=\sum_{k=n+1}^{an}\frac1k=\frac1n\sum_{k=n+1}^{an}\frac1{k/n}. $$ This last is a Riemann sum converging to $\int_1^a\frac1x\,dx=\log(a)$. The unblocked sequence $(s_n)$ of partial sums also converges to this limit, since the blocks defined by $(s_{an})$ are of fixed size and the terms in $(s_n)$ tend to zero.

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@Marek,

For me the hint was in http://oeis.org/A097321, from which the numerators for $\log(3)$ are $1$, $1$, $-2$.

Now $\log(2)$ having $1$, $-1$

and $\log(3)$ having $1$, $1$, $-2$

suggests the pattern.

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    $\begingroup$ The logarithm series in the question were known to at least Jaume Oliver Lafont in the OEIS before I posted the question above. $\endgroup$ Feb 5, 2023 at 13:30
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    $\begingroup$ Lehmer wrote $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}+\frac{1-k}{k}+\frac{1}{k+1}+...+\frac{1}{2k-1}+\frac{1-k}{2k}+...=\log(k)$ in 1975 matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf $\endgroup$ Feb 5, 2023 at 13:53
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For computing $log(\frac{p}{q})$ we can take $p$ positive terms from the harmonic series and $n$ negative ones at each step.

$$ \log\left(\frac{p}{q}\right)=\sum_{i=0}^\infty \left(\sum_{j=pi+1}^{p(i+1)}\frac{1}{j}-\sum_{k=qi+1}^{q(i+1)}\frac{1}{k}\right) $$

Sequence https://oeis.org/A166871 in the OEIS illustrates case $\frac{p}{q}=\frac{3}{2}$

This generalizes by using sequences as summation limits: https://math.stackexchange.com/a/1609512/134791

The expression for $\log\left(\frac{p}{q}\right)$ can also be written as a difference of summations of complex exponentials.

$$\log\left(\frac{p}{q}\right) = \sum_{n=1}^\infty \frac{1}{n}\left( \sum_{k=1}^{p-1}e^\frac{2\pi i kn}{p} -\sum_{k=1}^{q-1}e^\frac{2\pi i kn}{q} \right)$$

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