級数

${\displaystyle 1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots }$

${\displaystyle =[x-\frac{x^4}{4}+\frac{x^7}{7}-\frac{x^{10}}{10}+\cdots {]}_0^1}$

${\displaystyle ={\int }_0^1(1-x^3+x^6-x^9+\cdots )dx}$

${\displaystyle ={\int }_0^1\frac{1}{1+x^3}dx}$

${\displaystyle ={\int }_0^1\frac{1}{(x+1)(x^2-x+1)}dx}$

${\displaystyle =\frac{1}{3}{\int }_0^1\frac{1}{x+1}dx-\frac{1}{6}{\int }_0^1\frac{2x-1}{x^2-x+1}dx+\frac{1}{2}{\int }_0^1\frac{1}{x^2-x+1}dx}$

${\displaystyle =[\frac{1}{3}\ln (x+1)-\frac{1}{6}\ln (x^2-x+1)+\frac{1}{\sqrt{3}}\arctan \frac{2x-1}{\sqrt{3}}{]}_0^1}$

${\displaystyle =\frac{\sqrt{3}\pi }{9}+\frac{1}{3}\ln 2}$

3行前の三項目の積分は分母を平方完成すればいいです.

よって答えは${\displaystyle \frac{\sqrt{3}\pi }{9}+\frac{1}{3}\ln 2}$です.

最後まで読んでくれてありがとうございます.