How does the infinite series below simplify to that integral?
$$1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots=\int_0^1\frac{dx}{1+x^3}$$
I thought of simplifying the series to the sum to infinity of $\frac{1}{6n-5} - \frac{1}{6n-2}$, but this did not help.
$$\int_{0}^{1}{\frac{dx}{1-(-x)^3}=\int_{0}^{1}{\sum_{n=0}^{\infty}{(-x)}^{3n}}}dx=\sum_{n=0}^{\infty}{(-1)^{3n}\int_{0}^{1}{x^{3n}}dx}$$ $$=\sum_{n=0}^{\infty}{\frac{(-1)^{3n}}{3n+1}}= 1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+\cdots $$
for $x$ real, $n\geq 0$ integer \begin{align}\frac{1}{1+x^3}&=\frac{1-(-x^3)^{n+1}}{1-(-x^3)}+\frac{(-x^3)^{n+1}}{1-(-x^3)}\\ &=\frac{1-(-x^3)^{n+1}}{1-(-x^3)}+\frac{(-x^3)^{n+1}}{1+x^3}\\ \end{align}
For $x\neq 1$, $n\geq 0$ integer, \begin{align}\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x}\end{align}
Therefore, \begin{align}\int_0^1 \frac{1}{1+x^3}\,dx&=\int_0^1 \left(\sum_{k=0}^n (-x^3)^k\right)\,dx+\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\\ &=\sum_{k=0}^n \left(\int_0^1 (-x^3)^k\,dx\right)+\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\\ &=\sum_{k=0}^n \frac{(-1)^k}{3k+1}+\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\\ \end{align}
For $x\in[0;1],n\geq 0$, integer, \begin{align}\frac{x^{3(n+1)}}{1+x^3}\leq x^{3(n+1)}\end{align} and, \begin{align}\int_0^1 x^{3(n+1)}\,dx=\frac{1}{3n+4}\end{align} Therefore, \begin{align}\left|\int_0^1 \frac{(-x^3)^{n+1}}{1+x^3}\,dx\right|\leq \frac{1}{3n+4}\end{align} \begin{align}\left|\int_0^1 \frac{1}{1+x^3}\,dx-\sum_{k=0}^n \frac{(-1)^k}{3k+1}\right|\leq \frac{1}{3n+4}\end{align} Therefore, \begin{align}\boxed{\int_0^1 \frac{1}{1+x^3}\,dx=\sum_{k=0}^\infty \frac{(-1)^k}{3k+1}}\end{align}
If $\lvert x\rvert<1$, let$$f(x)=\sum_{n=0}^\infty\frac{x^{3n+1}}{3n+1}.$$Then $$f'(x)=\sum_{n=0}^\infty x^{3n}=\frac1{1-x^3}.$$Therefore\begin{align}1-\frac14+\frac17-\frac1{10}+\cdots&=\lim_{x\to1}f(x)\\&=\int_0^1f'(x)\,\mathrm dx\\&=\int_0^1\frac1{1-x^3}\,\mathrm dx.\end{align}
