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Question Number 185038 by Mastermind last updated on 16/Jan/23

$$\mathrm{Find}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{for}\mathrm{which}$$$$\mathrm{the}\mathrm{series}\frac{\mathrm{x}}{27}+\frac{{\mathrm{x}}^2}{125}+...+\frac{{\mathrm{x}}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^3}+...$$$$\mathrm{is}\mathrm{absolutely}\mathrm{convergent}.$$$$$$$$$$$$\mathrm{Help}!$$

Answered by FelipeLz last updated on 16/Jan/23

$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{{(2n+1)}^3}$$$$\underset{n\to \mathrm{\infty }}{\lim }\mid \frac{\frac{x^{n+1}}{{[2(n+1)+1]}^3}}{\frac{x^n}{{(2n+1)}^3}}\mid <1$$$$\underset{n\to \mathrm{\infty }}{\lim }\mid x\times {\left(\frac{2n+1}{2n+3}\right)}^3\mid <1$$$$\underset{n\to \mathrm{\infty }}{\lim }\mid x\mid \times \underset{n\to \mathrm{\infty }}{\lim }\mid {\left(\frac{2n+1}{2n+3}\times \frac{n^{-1}}{n^{-1}}\right)}^3\mid <1$$$$\mid x\mid \times \underset{n\to \mathrm{\infty }}{\lim }\mid {\left(\frac{2+n^{-1}}{2+3n^{-1}}\right)}^3\mid <1$$$$\mid x\mid <1$$

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