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Question Number 204544 by hardmath last updated on 21/Feb/24

$$\mathrm{Prove}\mathrm{that}:$$$$\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{{\mathrm{n}}^3}<\frac{1}{12}$$

Answered by mr W last updated on 21/Feb/24

$$\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}$$$$=(1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}+\frac{1}{{(n+1)}^3}+...)-1-\frac{1}{2^3}-(\frac{1}{{(n+1)}^3}+...)$$$$<(1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}+\frac{1}{{(n+1)}^3}+...)-1-\frac{1}{2^3}$$$$=\zeta \left(3\right)-\frac{9}{8}<1.2021-\frac{9}{8}<\frac{1}{12}$$

Commented by hardmath last updated on 23/Feb/24

$$\mathrm{Perfect}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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