How to evaluate $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{12n+1} $? First of all, I am going to convert this series into a definite integral.
For $|x|<1,$ we have
$\displaystyle \quad \frac{1}{1+x^{12}}=\displaystyle \sum_{n=0}^{\infty}\left(-x^{12}\right)^{n}=\displaystyle \sum_{n=0}^{\infty}(-1)^{n} x^{12 n} \tag*{} $
Integrating both sides from $x=0$ to $1$, we can relate the target series and the definite integral.
$$\displaystyle \int_{0}^{1} \frac{d x}{1+x^{12}} =\displaystyle \sum_{n=0}^{\infty} \int_{0}^{1}(-1)^{n} x^{12 n} d x 
\displaystyle =\sum_{n=0}^{\infty}\left[(-1)^{n} \frac{x^{12 n+1}}{12n+1}\right]_{0}^{1} \displaystyle =\sum_{n=0}^{\infty} \frac{(-1)^{n}}{12 n+1}$$
By my post,
$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}}{12 n+1}$
$\displaystyle \\ \displaystyle  =\frac{1}{24 \sqrt{2}}\left[2 \ln \left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)+(1+\sqrt{3}) \ln \left(\frac{2 \sqrt{2}+\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}-1}\right)\right.\\$
$\displaystyle \qquad\qquad \qquad\qquad \left.+(\sqrt{3}-1) \ln \left(\frac{2 \sqrt{2}+\sqrt{3}-1}{2 \sqrt{2}-\sqrt{3}+1}\right)+2(1+\sqrt{3}) \pi\right] \\$
(checked by Wolframalpha)
My question
How to generalize the result for
$$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k n+1}$$
where $n\in \mathbb N$?

Similar to the original integral, using the same technique yields
$\displaystyle \quad \frac{1}{1+x^{n}}=\displaystyle \sum_{k=0}^{\infty}\left(-x^{n}\right)^{k}=\displaystyle \sum_{k=0}^{\infty}(-1)^{k} x^{nk} ,\tag*{}
$
we have
Integrating both sides from $x=0$ to $1$, we can relate the target series and the definite integral.
$$\displaystyle \int_{0}^{1} \frac{d x}{1+x^{n}} =\displaystyle \sum_{k=0}^{\infty} \int_{0}^{1}(-1)^{k} x^{nk} d x 
\displaystyle =\sum_{k=0}^{\infty}\left[(-1)^{k} \frac{x^{nk+1}}{nk+1}\right]_{0}^{1} \displaystyle =\sum_{k=0}^{\infty} \frac{(-1)^{k}}{nk+1}$$
By the Dr Brian Sittinger’s post,
A. when $n=2m$ is even, we obtain the series$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k n+1}\\=\boxed{\frac{1}{2 m} \sum_{k=0}^{m-1}\left[\frac{(2 k+1) \pi}{2 m} \sin \left(\frac{(2 k+1) \pi}{2 m}\right)-\cos \left(\frac{(2 k+1) \pi}{2 m}\right) \ln \left(2-2 \cos \left(\frac{(2 k+1) \pi}{2 m}\right)\right)\right]}
$$
B. when $n=2m+1$ is odd, we obtain the series$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k n+1}\\ =\boxed{\frac{\ln 2}{2 m+1}+\frac{1}{2 m+1} \sum_{k=0}^{m-1}\left[\frac{(2 k+1) \pi}{2 m+1} \sin \left(\frac{(2 k+1) \pi}{2 m+1}\right)-\cos \left(\frac{(2 k+1) \pi}{2 m+1}\right) \ln \left(2-2 \cos \left(\frac{(2 k+1) \pi}{2 m+1}\right)\right)\\+2 \sin \left(\frac{(2 k+1) \pi}{2 m+1}\right) \cdot \arctan \left(\cot \left(\frac{(2 k+1) \pi}{2 m+1}\right)\right)\right]}
$$
