Question Number 165829 by HongKing last updated on 09/Feb/22
$$\mathrm{Compare}\:\mathrm{it}: \\ $$$$\mathrm{p}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} }\:\:\mathrm{and}\:\:\mathrm{q}\:=\:\mathrm{0},\mathrm{99} \\ $$$$\left(\mathrm{a}\right)\mathrm{p}=\mathrm{q}\:\:\left(\mathrm{b}\right)\mathrm{p}<\mathrm{q}\:\:\left(\mathrm{c}\right)\mathrm{p}>\mathrm{q}\:\:\left(\mathrm{d}\right)\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{e}\right)\:\sqrt{\mathrm{q}}\:=\:\sqrt{\mathrm{p}}\:-\:\mathrm{2} \\ $$
Answered by MJS_new last updated on 09/Feb/22
$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{weird}\:\mathrm{question} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}<\frac{\left(\frac{\mathrm{22}}{\mathrm{7}}\right)^{\mathrm{2}} }{\mathrm{6}}=\frac{\mathrm{242}}{\mathrm{147}}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}<\frac{\mathrm{95}}{\mathrm{147}}<\frac{\mathrm{98}}{\mathrm{147}}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{2}} {\overset{\mathrm{100}} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:<\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$$\Rightarrow\:{p}<{q} \\ $$