## Irrational logs and the harmonic series

Consider the series $$S_f = \sum_{x=1}^\infty \frac{f}{x^2+fx}.$$ Goldbach showed that, for integers $f \ge 1$, $$S_f = 1 + \frac12 + \frac13 + \ldots + \frac1f$$ (this follows easily by writing $S_f$ as a telescoping series). Thus $S_f$ is rational for all natural numbers $f \ge 1$. Goldbach claimed that, for all nonintegral (rational) numbers $f$, the sum $S_f$ would be irrational.

Euler showed, by using the substitution $$\frac1k = \int_0^1 x^{k-1} dx,$$ that $$S_f = \int_0^1 \frac{1-x^f}{1-x} dx.$$ He evaluated this integral for $f = \frac12$ and found that $S_{1/2} = 2(1 - \ln 2)$ (this also follows easily from Goldbach's series for $S_f$). Thus Goldbach's claim holds for all $f \equiv \frac12 \bmod 1$ since $S_{f+1} = S_f + \frac1{f+1}$.

Here are my questions:

1. The irrationality of $\ln 2$ was established by Lambert, who proved that $e^r$ is irrational for all rational numbers $r \ne 0$. Are there any (simple) direct proofs?

2. Has Goldbach's claim about the irrationality of $S_f$ for nonintegral rational values of $f$ been settled in other cases?

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I'm suspicious there are (simple) direct proofs for ln2 being irrational (the standard argument for proving e^r is irrational seems simple enough). I did find this cute proof however: math.sc.edu/~filaseta/gradcourses/Math785/… – Alex R. Feb 7 2010 at 17:58
Dear Franz, the function S_f is equal to P(f+1)-P(1), where P(f) is the digamma function (the logarithmic derivative of the Gamma function). There is a formula by Gauss (Gauss' digamma theorem) which then gives an explicit formula for P(p/q)-P(1) in terms of a linear combination (with algebraic coefficients) of logarithms of algebraic numbers. Here (of course) one is using that log(-1) = pi*i. Irrationality (in fact, transcendence) then follows (with a little work) from Baker's theorem on linear forms in logarithms. – Lavender Honey Feb 10 2010 at 19:31
It appears that FC answered the second question, but here is another concrete example. If we let f=1/4, then the integral becomes 4-(pi/2)-ln(8). In general, the linear combination mentioned by FC will always have a nonzero algebraic coefficient on pi, except when p/q=1/2 [and that case was dealt with already]. – Pace Nielsen Feb 10 2010 at 20:41
@FC: You should write this as an answer, and claim the bounty. – Theo Johnson-Freyd Feb 10 2010 at 21:07
If the bounty consisted of a bottle of Bordeaux, then it might be something worth claiming... – Lavender Honey Feb 14 2010 at 18:35