# 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:

$$\begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$
Is it true that \displaystyle \log(n)=\sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}$$\;? - And why do you expect this to hold (e.g. you computed the partial sums up to a big number and they are close to the \logs...)? – Marek Jun 19 '11 at 23:36 Why's this tagged as number-theory? – George Lowther Jun 20 '11 at 1:46 @Marek & @George Lowther: Perhaps not an answer to your questions, but this recurrence here: list.seqfan.eu/pipermail/seqfan/2011-June/014999.html , led me to the values of the Mangoldt function here: list.seqfan.eu/pipermail/seqfan/2011-June/015006.html , which in turn led me to the series above. – Mats Granvik Jun 20 '11 at 8:07 right, thanks. I'm glad it wasn't just a guess because to me the limit was quite unapparent (although I admit I am ignorant and series like these might be well-known). – Marek Jun 20 '11 at 8:59 A while ago, after I asked this question, I noticed that these logarithm series have been known to Jaume Oliver Lafont in the Oeis: oeis.org/wiki/User:Jaume_Oliver_Lafont – Mats Granvik Aug 29 '11 at 18:34 ## 5 Answers You can write T(n,k)=1-n1_{\{n\mid k\}}. Then, for \vert x\vert < 1 look at the power series$$ \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. $$- Very nice derivation from basic principles. – André Nicolas Jun 20 '11 at 1:43 Yes. You can get the sums by differentiating the digamma function repeatedly. There is a good deal of information about the resulting polygamma functions, including series expressions, here. Your matrix version is a lot more visually arresting than the usual Dirac delta function formulation! - I'd still like to know how one comes up with such series and a conjecture of what they should converge to. Any clues? – Marek Jun 20 '11 at 1:07 @Marek: (This is only speculation.) There is a long history of evaluation of \Gamma, \Gamma'/\Gamma, and their derivatives at special points. So the answers (\log n) may have come before the questions (series). – André Nicolas Jun 20 '11 at 1:51 @AndréNicolas When you say resulting polygamma functions, what do you mean? I am trying to find a relationship between n/LambertW(n)-1 and Stirling numbers of the second kind. In the derivative of the explicit formula for Stirling numbers of the second kind I get the polygamma function, according to Mathematica. How do you find the polygamma function in relation to this question about logarithms? – Mats Granvik Oct 18 '13 at 12:09 After some experimenting, I find that$$\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)$$where \text{Li}_s\left(x^2\right) is the PolyLog function and \Phi \left(x^2,s,\frac{1}{2}\right) is the LerchPhi function. Is this what you meant in your answer above? – Mats Granvik Oct 18 '13 at 12:16 @Marek, For me the hint was in http://oeis.org/A097321, from which the numerators for log(3) are 1,1,-2. Now log(2) having 1, -1 and log(3) having 1, 1, -2 suggests the pattern. - The formula seem to be extendable to fractional arguments of the log. The key is to rewrite the formula for \log(x) as difference of two sums but to a common limit. So we can write$$ \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 $$suggests to look into wikipedia to see, who had noticed that first... ;-) and it's nice to see the identity of the integral-definition and the simple reformulation and generalization of your surprising patterns. - For computing log(\frac{p}{q}) we can take p positive terms from the harmonic series and n negative ones at each step.$$ \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: http://math.stackexchange.com/a/1609512/134791

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