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Question Number 204545 by hardmath last updated on 21/Feb/24

$$\mathrm{If}$$$$\mathrm{a}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}$$$$\mathrm{b}=0,99$$$$\mathrm{Prove}\mathrm{that}:$$$$\mathrm{a}<\mathrm{b}$$

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

$$a=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}$$$$=(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}+\frac{1}{{101}^2}+...)-1-(\frac{1}{{101}^2}+...)$$$$<(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}+\frac{1}{{101}^2}+...)-1$$$$=\frac{{\pi }^2}{6}-1<0.99$$

Answered by witcher3 last updated on 21/Feb/24

$${\mathrm{k}}^2⩾\mathrm{k}(\mathrm{k}-1);\mathrm{\forall }\mathrm{k}⩾2$$$$\Rightarrow \frac{1}{{\mathrm{k}}^2}⩽\frac{1}{\mathrm{k}(\mathrm{k}-1)}\Rightarrow \mathrm{a}=\underset{\mathrm{k}=2}{\sum ^{100}}\frac{1}{{\mathrm{k}}^2}⩽\underset{\mathrm{k}=2}{\sum ^{100}}\frac{1}{\mathrm{k}(\mathrm{k}-1)}=\underset{\mathrm{k}=2}{\sum ^{100}}(\frac{1}{\mathrm{k}-1}-\frac{1}{\mathrm{k}})=1-\frac{1}{100}=0.99$$$$\mathrm{a}<\mathrm{b}$$

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

$$greatapproach!$$$$butwiththismethodwecannotprove$$$$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}<\frac{3}{4}$$

Commented by witcher3 last updated on 22/Feb/24

$$\mathrm{exactly}\mathrm{same}\mathrm{aproch}\mathrm{will}\mathrm{not}\mathrm{Give}\mathrm{this}$$$$\mathrm{but}\mathrm{i}\mathrm{think}\mathrm{That}\mathrm{its}\mathrm{to}\mathrm{prove}<\frac{3}{4}$$$$\mathrm{using}\mathrm{interpelation}\mathrm{of}\mathrm{lagrange}$$$$\mathrm{approching}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2};\mathrm{withe}\mathrm{polynome}$$

Commented by mr W last updated on 22/Feb/24

$$thanksalot!$$

Commented by hardmath last updated on 23/Feb/24

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professors}$$

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