Question Number 71664 by mathmax by abdo last updated on 18/Oct/19
$${find}\:{nature}\:{of}\:{the}\:{sequence}\:{U}_{{n}} =\frac{\mathrm{1}}{{n}}\left(\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\right)^{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 18/Oct/19
$${we}\:{have}\:\sum_{{i}=\mathrm{1}} ^{{n}} \:{x}_{{i}} ^{\mathrm{2}} \geqslant\frac{\mathrm{1}}{{n}}\left(\sum_{{i}=\mathrm{1}} ^{{n}} {x}_{{i}} \right)^{\mathrm{2}} \:\:\:\left({result}\:{proved}\right) \\ $$$${let}\:{x}_{{i}} =\frac{\mathrm{1}}{{i}}\:\Rightarrow\sum_{{i}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{i}^{\mathrm{2}} }\:\geqslant\frac{\mathrm{1}}{{n}}\left(\sum_{{i}=\mathrm{1}} ^{{i}} \:\frac{\mathrm{1}}{{i}}\right)^{\mathrm{2}} \:\Rightarrow\mathrm{0}<{U}_{{n}} \leqslant\sum_{{i}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{i}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{0}\leqslant{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \leqslant\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\:{the}\:{sequenc}\:{U}_{{n}} {converges}. \\ $$$${another}\:{way}\:\:{we}\:{have}\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}\:={H}_{{n}} ={ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow \\ $$$$\Rightarrow{H}_{{n}} \sim{ln}\left({n}\right)+\gamma\:\Rightarrow{H}_{{n}} ^{\mathrm{2}} \sim{ln}^{\mathrm{2}} \left({n}\right)+\mathrm{2}\gamma{ln}\left({n}\right)+\gamma^{\mathrm{2}} \:\:\sim{ln}^{\mathrm{2}} \left({n}\right)\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${U}_{{n}} \sim\frac{{ln}^{\mathrm{2}} \left({n}\right)}{{n}}\:\:{we}\:{have}\:{lim}_{{n}\rightarrow+\infty} \:\:\:\frac{{ln}^{\mathrm{2}} \left({e}^{{n}} \right)}{{e}^{{n}} }\:={lim}_{{n}\rightarrow+\infty} \:\:{n}^{\mathrm{2}} \:{e}^{−{n}} =\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{U}_{{n}} =\mathrm{0} \\ $$
Answered by mind is power last updated on 18/Oct/19
$$\forall\mathrm{n}\in\mathbb{N}^{\ast} \:\:\:\mathrm{we}\:\mathrm{have}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \frac{\mathrm{1}}{\mathrm{k}}\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{Un}\geqslant\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\Sigma\mathrm{U}_{\mathrm{n}} \geqslant\Sigma\frac{\mathrm{1}}{\mathrm{n}}\rightarrow+\infty \\ $$
Commented by mathmax by abdo last updated on 18/Oct/19
$${sir}\:{the}\:{question}\:{is}\:{find}\:{the}\:{nature}\:{of}\:{the}\:{sequence}\:{not}\:{the}\:{serie}… \\ $$
Commented by mind is power last updated on 18/Oct/19
$$\mathrm{ok}\:\mathrm{sorry}\: \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{sequence}\: \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}\backsim\mathrm{ln}\left(\mathrm{n}\right) \\ $$$$\Rightarrow\mathrm{Un}\backsim\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{n}\right)}{\mathrm{n}}\rightarrow\mathrm{0} \\ $$$$\mathrm{un}\rightarrow\mathrm{0} \\ $$