# Do these series converge to logarithms?

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}$$\;? • And why do you expect this to hold (e.g. you computed the partial sums up to a big number and they are close to the \logs...)? Jun 19, 2011 at 23:36 • Why's this tagged as number-theory? Jun 20, 2011 at 1:46 • @Marek & @George Lowther: Perhaps not an answer to your questions, but this recurrence here: list.seqfan.eu/pipermail/seqfan/2011-June/014999.html , led me to the values of the Mangoldt function here: list.seqfan.eu/pipermail/seqfan/2011-June/015006.html , which in turn led me to the series above. Jun 20, 2011 at 8:07 • We can see the formula for log(k) in page 136 by Lehmer (1975) matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf Feb 4, 2016 at 11:34 • The periodicity of T(n,k) suggests Fourier analysis, and the result is remarkable: since it is a zero-mean delta train, it has all components but the constant.$$\log(n) = \sum_{k=1}^\infty \frac{1}{k} \sum_{j=1}^{n-1} e^\frac{2 \pi ijk}{n} $$Feb 5, 2023 at 12:19 ## 6 Answers You can write T(n,k)=1-n1_{\{n\mid k\}}. Then, for \vert x\vert < 1 look at the power series$$ \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. $$• Very nice derivation from basic principles. Jun 20, 2011 at 1:43 Yes. You can get the sums by differentiating the digamma function repeatedly. There is a good deal of information about the resulting polygamma functions, including series expressions, here. Your matrix version is a lot more visually arresting than the usual Dirac delta function formulation! • I'd still like to know how one comes up with such series and a conjecture of what they should converge to. Any clues? Jun 20, 2011 at 1:07 • @Marek: (This is only speculation.) There is a long history of evaluation of \Gamma, \Gamma'/\Gamma, and their derivatives at special points. So the answers (\log n) may have come before the questions (series). Jun 20, 2011 at 1:51 • @AndréNicolas When you say resulting polygamma functions, what do you mean? I am trying to find a relationship between n/LambertW(n)-1 and Stirling numbers of the second kind. In the derivative of the explicit formula for Stirling numbers of the second kind I get the polygamma function, according to Mathematica. How do you find the polygamma function in relation to this question about logarithms? Oct 18, 2013 at 12:09 • After some experimenting, I find that$$\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)$$where \text{Li}_s\left(x^2\right) is the PolyLog function and \Phi \left(x^2,s,\frac{1}{2}\right) is the LerchPhi function. Is this what you meant in your answer above? Oct 18, 2013 at 12:16 The formula seem to be extendable to fractional arguments of the log. The key is to rewrite the formula for \log(x) as difference of two sums but to a common limit. So we can write$$ \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. 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. @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. • The logarithm series in the question were known to at least Jaume Oliver Lafont in the OEIS before I posted the question above. Feb 5, 2023 at 13:30 • 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 Feb 5, 2023 at 13:53

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.

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