There is a solution: Evaluating $\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$
But my question is why $ \sum_{n=0}^\infty (-x)^{3n} = \frac{1}{1 + x^{3}} $. Isn't it work only for $ |(-x)^3|< 1 $ ?
2026-03-26 01:14:30.1774487670
Question about evaluating $\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$ (detail)
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I guess the question is for this equation: $$ \int_{0}^{1}\sum_{n\geq 0}(-x)^{3n}\,dx =\int_{0}^{1}\frac{dx}{1+x^3} $$ The two integrands agree almost everywhere, so the integrals are equal. The only exception is (as noted) the value $x=1$.
(That is the Lebesgue version of the answer. There is also a Riemann version, where we (potintially) consider the left-hand integral to be improper at the endpoint $x=1$.)