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Question Number 165829 by HongKing last updated on 09/Feb/22

$$\mathrm{Compare}\mathrm{it}:$$ $$\mathrm{p}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{{100}^2}\mathrm{and}\mathrm{q}=0,99$$ $$\left(\mathrm{a}\right)\mathrm{p}=\mathrm{q}\left(\mathrm{b}\right)\mathrm{p}<\mathrm{q}\left(\mathrm{c}\right)\mathrm{p}>\mathrm{q}\left(\mathrm{d}\right){\mathrm{p}}^2+{\mathrm{q}}^2=0$$ $$\left(\mathrm{e}\right)\sqrt{\mathrm{q}}=\sqrt{\mathrm{p}}-2$$

Answered by MJS_new last updated on 09/Feb/22

$$\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{weird}\mathrm{question}$$ $$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}<\frac{{\left(\frac{22}{7}\right)}^2}{6}=\frac{242}{147}\Rightarrow$$ $$\Rightarrow \underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}-1<\frac{95}{147}<\frac{98}{147}=\frac{2}{3}\Rightarrow$$ $$\Rightarrow \underset{n=2}{\sum ^{100}}\frac{1}{n^2}<\frac{2}{3}\Rightarrow$$ $$\Rightarrow p<q$$

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