Harmonic series with sign alternates every $n$ terms.

Question

Let $A(1)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$

Let $A(2)=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\dots$

Let $A(3)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}+\dots$

In general, for $n \in \mathbb{N}$, let $A(n)$ be the same series as the above with sign alternates every $n$ terms.

I know that $A(1)=\log(2)$, and $A(2)=\frac{1}{4}(\pi+\log(4))$.

How can we evaluate $A(n)$ for $n \in \mathbb{N_{\ge3}}$?

Answer 1

$A(3) = \lim_{m\to\infty} K_m$ where $$ K_m = \sum_{k=1}^m\frac{1}{6k-5} +\sum_{k=1}^m\frac{1}{6k-4} +\sum_{k=1}^m\frac{1}{6k-3} -\sum_{k=1}^m\frac{1}{6k-2} -\sum_{k=1}^m\frac{1}{6k-1} -\sum_{k=1}^m\frac{1}{6k} $$ Then use estimates like $$ \sum_{k=1}^m\frac{1}{6k-a} = \frac{1}{6}\left(\log m-\psi\left(1-\frac{a}{6}\right)\right)+O(1/m) $$ with the Gauss digamma theorem for $\psi$ of rational number. Result: $$ A(3) = \frac{\log 2}{3}+\frac{2\pi}{3\sqrt{3}} . $$

Answer 2

$A_n$ can be expressed as $$ A_n=\sum_{k=1}^n \sum_{u=0}^\infty \frac{(-1)^u}{nu+k}. $$ Now, from the Taylor series of $\frac{1}{1+x}$ for $|x|<1$ we have $$ { \frac{1}{1+x}=\sum_{u=0}^\infty (-1)^ux^u\implies \\ \frac{1}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{nu}\implies \\ \frac{x^{k-1}}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{nu+k-1}\implies \\ \int_0^1\frac{x^{k-1}}{1+x^n}dx=\sum_{u=0}^\infty \frac{(-1)^u}{nu+k} } $$ hence $$ A_n{=\sum_{k=1}^n \int_0^1\frac{x^{k-1}}{1+x^n}dx \\= \int_0^1\sum_{k=1}^n\frac{x^{k-1}}{1+x^n}dx \\= \int_0^1\frac{1-x^n}{1+x^n}\frac{1}{1-x}dx \\= \int_0^1\sum_{k=1}^n \frac{2u_k}{u_k-1}\frac{1}{x-u_k} dx \\= \sum_{k=1}^n\frac{2u_k}{u_k-1}\int_0^1 \frac{1}{x-u_k} dx \\= \sum_{k=1}^n\frac{2u_k}{u_k-1}[\ln(1-u_k)-\ln(-u_k)] } $$ where $u_k=\exp(i\pi\frac{2k+1}{n})$ is the $k$th root of $-1$ and $\ln$ is the main branch of complex logarithm.

Additional Remarks

The reason for interchanging integrals and summations, is the existence and boundedness of the integral values and the boundedness of summation scope.