$\displaystyle\sum^{+\infty}_{n=1}\dfrac{(-1)^{n-1}}{3n-2}$ By expanding $\dfrac{1}{1+x^3}$. I get a wrong answer$\dfrac{\ln2}{3}+\dfrac{\sqrt{3}\pi}{18}$.Why?
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Related: https://math.stackexchange.com/questions/1592898/ – Watson Oct 30 '18 at 13:50
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How should we know? By integrating $\frac{x}{1+x^3}$ over $(0,1)$ you should get the right answer. – Jack D'Aurizio Oct 30 '18 at 13:52
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Hint:
$$xS=x\sum_{n=0}^\infty\dfrac {x^{3n-3}}{3n-2}=\int\sum_{n=0}^\infty x^{3n-3}dx=?$$
Here $x^3=-1$
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