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Let define, for $k \ge 1$ : $$ f(k) = \sum_\limits{n=0}^{\infty} \frac{(-1)^n}{kn+1}. $$

It is well-known that $f(1) = \ln(2), f(2) = \pi/4$. Some computations on WolframAlpha led me to $f(3) =1/9 (\sqrt3 \pi+\ln(8))$, $f(4) = (\pi+2 \ln(1+\sqrt2))/(4 \sqrt2)$ and also (if I'm not mistaken) : $$ f(5) = 1/b \cdot \Big(\frac{8\sqrt2}{\sqrt a} \;\pi \;-\; 6 (\sqrt5 - 1)\ln(2) \;+\; 2 (3-\sqrt5)\ln(\sqrt 5 + 1)\\ - 4 \ln(\sqrt5 - 1) \;+\; (\sqrt5 - 5)\ln\Big( \frac{\bar a}{a} \Big) \Big)$$

with $a = 5+\sqrt5, \bar a = 5-\sqrt5,b=20(\sqrt 5 - 1)$.

Then, I would like to ask the following : is it true that (for general $k \geq 1$) $f(k) \in \overline{ \mathbb Q} (A)$ where $A = \{ \pi \} \cup \{ \ln(x) \mid x \in \overline{ \mathbb Q} \cap \mathbb R \}$, as it seems to be the case for small values of $k$ ? Are there some available results on these series?

I looked at some special functions : this result on the digamma function is related to my question. I don't know if it is possible to use this result in order to compute $f(k)$.

Any comment or answer would be appreciated!

Watson
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1 Answers1

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We have: $$ f(k) = \sum_{n\geq 0}(-1)^n \int_{0}^{1} x^{kn}\,dx = \int_{0}^{1}\frac{dx}{1+x^k} $$ and the last integral can be easily computed through the residue theorem, since $\frac{1}{1+x^k}$ has simple poles at $\zeta_j = \exp\left(\frac{\pi i}{k}(2j-1)\right)$ for $j=1,2,\ldots,k$ with residues given by: $$ \text{Res}\left(\frac{1}{1+x^k},x=\zeta_j\right) = \frac{\zeta_j}{k \zeta_j^{k}}=-\frac{\zeta_j}{k}.$$ Since $\int_{0}^{1}\frac{dx}{x-\zeta_j}\,dx = \log\left(1-\frac{1}{\zeta_j}\right)$, we get:

$$ f(k) = -\frac{1}{k}\sum_{j=1}^{k}\zeta_j \log\left(1-\frac{1}{\zeta_j}\right) $$

and that can be further simplified by coupling terms related with conjugated roots, leading to the $\log\cos$ contributes mentioned in the comments.

Jack D'Aurizio
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