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Question Number 40898 by abdo.msup.com last updated on 28/Jul/18
$${let}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{{n}−{k}}{{n}−{k}+\mathrm{1}} \\ $$$${find}\:{a}\:{equivalent}\:{of}\:{u}_{{n}} \left({n}\rightarrow+\infty\right) \\ $$
Commented by maxmathsup by imad last updated on 29/Jul/18
$${thanks}\:{i}\:{understand}.... \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18
$${T}_{{k}} =\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}−{k}} \\ $$$${T}_{\mathrm{1}} =\mathrm{1}−\frac{\mathrm{1}}{{n}} \\ $$$${T}_{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{1}}{{n}−\mathrm{1}} \\ $$$$... \\ $$$$... \\ $$$${T}_{{n}−\mathrm{1}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:{u}_{{n}} =\left({n}−\mathrm{1}\right)−\left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{{n}}\right\}\:{to}\:{find} \\ $$$$ \\ $$$$\:\:{u}_{{n}} =\left({n}−\mathrm{1}\right)−{s} \\ $$$${value}\:{of}\:{u}_{{n}} \:{n}\rightarrow\infty \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}>\frac{\mathrm{1}}{\mathrm{2}}>\frac{\mathrm{1}}{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}>\frac{\mathrm{1}}{\mathrm{3}}>\frac{\mathrm{1}}{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}>\frac{\mathrm{1}}{\mathrm{4}}>\frac{\mathrm{1}}{{n}} \\ $$$$..... \\ $$$$...... \\ $$$${so}\:\:\:{adding}\:\:{n}−\mathrm{1}>{s}>\frac{{n}−\mathrm{1}}{{n}} \\ $$$$\:\:\:\:−\left({n}−\mathrm{1}\right)<−{s}<−\left(\frac{{n}−\mathrm{1}}{{n}}\right) \\ $$$$\left({n}−\mathrm{1}\right)−\left({n}−\mathrm{1}\right)<\left({n}−\mathrm{1}\right)−{s}<\left({n}−\mathrm{1}\right)−\left(\frac{{n}−\mathrm{1}}{{n}}\right) \\ $$$$\mathrm{0}<{u}_{{n}} <{n}−\mathrm{1}−\mathrm{1}+\frac{\mathrm{1}}{{n}} \\ $$$$\mathrm{0}<{u}_{{n}} <{n}+\frac{\mathrm{1}}{{n}}−\mathrm{2} \\ $$$$\mathrm{0}<{u}_{{n}} <\left(\sqrt{{n}}\:−\frac{\mathrm{1}}{\sqrt{{n}_{\:} }}\right)^{\mathrm{2}} \\ $$$${when}\:{n}\rightarrow\infty\:\:\:\:\:\:\infty>{u}_{{n}} >\mathrm{0} \\ $$$${pls}\:{check}.... \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by math khazana by abdo last updated on 29/Jul/18
$${we}\:{have}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{{n}−{k}+\mathrm{1}−\mathrm{1}}{{n}−{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{1}\right)−\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{n}−{k}+\mathrm{1}}\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{1}\right)={n}−\mathrm{1}\:{and} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{n}−{k}+\mathrm{1}}\:=_{{n}−{k}={i}} \:\:\:\sum_{{i}={n}−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}}{{i}+\mathrm{1}}\:=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{i}+\mathrm{1}} \\ $$$$=\sum_{{i}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{i}}\:={H}_{{n}} −\mathrm{1}\:\Rightarrow \\ $$$${u}_{{n}} ={n}−\mathrm{1}−{H}_{{n}} +\mathrm{1}\:={n}−{H}_{{n}} \:\:{but} \\ $$$${H}_{{n}} ={ln}\left({n}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:\Rightarrow{u}_{{n}} \:={n}−{ln}\left({n}\right)−\gamma+{o}\left(\frac{\mathrm{1}}{{n}}\right)\Rightarrow \\ $$$${u}_{{n}} \sim{n}−{ln}\left({n}\right)\left({n}\rightarrow+\infty\right) \\ $$
Commented by math khazana by abdo last updated on 29/Jul/18
$${H}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\: \\ $$