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Question Number 141640 by Willson last updated on 21/May/21

Answered by qaz last updated on 22/May/21

$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{16}^k(8k+n)}$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{16}^k}{\int }_0^1{x}^{8k+n-1}dx$$$$={\int }_0^1{x}^{n-1}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{x^{8k}}{{16}^k}dx$$$$={\int }_0^1\frac{x^{n-1}}{1-\frac{x^8}{16}}dx$$$$=16{\int }_0^1\frac{x^{n-1}}{16-x^8}dx$$$$=16\sqrt{2}{\int }_0^{1/\sqrt{2}}\frac{{\left(\sqrt{2}x\right)}^{n-1}}{16-{\left(\sqrt{2}x\right)}^8}dx$$$$={\left(\sqrt{2}\right)}^n{\int }_0^{1/\sqrt{2}}\frac{x^{n-1}}{1-x^8}dx$$

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