Question Number 159497 by naka3546 last updated on 17/Nov/21
$${Prove}\:\:{that}\:\: \\ $$$$\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{9}}\:+\:\ldots+\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:<\:\:\mathrm{2} \\ $$$${for}\:\:{n}\:\in\:\:\mathbb{N}\:\:. \\ $$
Commented by mr W last updated on 18/Nov/21
$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$<\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+…=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$<\frac{\mathrm{10}}{\mathrm{6}}<\frac{\mathrm{12}}{\mathrm{6}}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }<\mathrm{2} \\ $$
Commented by naka3546 last updated on 18/Nov/21
$${Thank}\:\:\:{you}\:\:,\:\:{sir}. \\ $$
Commented by SANOGO last updated on 18/Nov/21
$${prkw}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:{explique}\:{moi}\:{un}\:{peu} \\ $$
Commented by mr W last updated on 18/Nov/21
https://en.m.wikipedia.org/wiki/Basel_problem