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Question Number 212164 by mnjuly1970 last updated on 04/Oct/24

$$$$$$$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{3n+1}-\ln \left(\sqrt[3]{2}\right)=?$$$$$$$$$$

Answered by Berbere last updated on 04/Oct/24

$$=\sum _{n⩾0}{\int }_0^1{(-x^3)}^{n}dx-\frac{ln\left(2\right)}{3}={\int }_0^1\frac{1}{1+x^3}dx-\frac{ln\left(2\right)}{3}$$$$={\int }_0^1\frac{1}{(x+1)(x^2-x+1)}dx-\frac{ln\left(2\right)}{3}$$$$={\int }_0^1\frac{(x+1)(x-2)-(x^2-x+1)}{-3(x+1)(x^2-x+1)}dx-\frac{ln\left(2\right)}{3}$$$$={\int }_0^1\frac{x-2}{-3(x^2-x+1)}+\frac{1}{3(x+1)}dx-\frac{ln\left(2\right)}{3}$$$$=\frac{1}{3}{\int }_0^1\frac{2-x}{x^2-x+1)}dx=-\frac{1}{6}{\int }_0^1\frac{2x-1}{x^2-x+1}dx+\frac{1}{2}{\int }_0^1\frac{dx}{x^2-x+1}$$$${=\frac{1}{2}{\int }_0^1\frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}=\frac{1}{\sqrt{3}}{\tan }^{-1}(\frac{2x}{\sqrt{3}}-\frac{1}{\sqrt{3}})]}_0^1$$$$=\frac{2}{\sqrt{3}}.\frac{\pi }{6}=\frac{\pi }{3\sqrt{3}}$$

Commented by mnjuly1970 last updated on 04/Oct/24

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