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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: $T(n,k) = -(n-1)\; \text{ if }\; n|k, \;\text{ else } \;1,$


$$\begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$
Is it true that $\displaystyle \log(n)=\sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}$$\;$?

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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...)? – 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: , led me to the values of the Mangoldt function here: , 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: – Mats Granvik Aug 29 '11 at 18:34
up vote 20 down vote accepted

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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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!

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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


For me the hint was in, 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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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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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 in the OEIS illustrates case $\frac{p}{q}=\frac{3}{2}$

This generalizes by using sequences as summation limits:

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