What do we know about $\sum_\limits{n=0}^{\infty} \frac{(-1)^n}{kn+1}$?

Solution 1:

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.