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

Prove that: (1/3^3 ) + (1/4^3 ) + ... + (1/n^3 ) < (1/(12))

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

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

(1/3^3 )+(1/4^3 )+...+(1/n^3 ) =(1+(1/2^3 )+(1/3^3 )+(1/4^3 )+...+(1/n^3 )+(1/((n+1)^3 ))+...)−1−(1/2^3 )−((1/((n+1)^3 ))+...) <(1+(1/2^3 )+(1/3^3 )+(1/4^3 )+...+(1/n^3 )+(1/((n+1)^3 ))+...)−1−(1/2^3 ) =ζ(3)−(9/8)<1.2021−(9/8)<(1/(12))

$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+...+\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+...+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }+...\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }−\left(\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }+...\right) \\ $$$$<\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+...+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }+...\right)−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} } \\ $$$$=\zeta\left(\mathrm{3}\right)−\frac{\mathrm{9}}{\mathrm{8}}<\mathrm{1}.\mathrm{2021}−\frac{\mathrm{9}}{\mathrm{8}}<\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented by hardmath last updated on 23/Feb/24

Perfect dear professor thank you

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

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