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Question Number 40504 by prof Abdo imad last updated on 23/Jul/18

let u_n =Σ_(k=1) ^n (1/(n^2 +k)) find lim_(n→+∞) n{1−n u_n } .

$${let}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{k}} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} {n}\left\{\mathrm{1}−{n}\:{u}_{{n}} \right\}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18

u_n =(1/(n^2 +1))+(1/(n^2 +2))+...+(1/(n^2 +n)) (1/(n^2 +1))> (1/(n^2 +1))>(1/(n^2 +n)) (1/(n^2 +1))>(1/(n^2 +2))>(1/(n^2 +n)) ... ... (1/(n^2 +1))>(1/(n^2 +n))>(1/(n^2 +n)) T_n ←(n/(n^2 +1))>u_n >(n/(n^2 +n))→v_n lim_(n→∞) n{1−nT_n } lim_(n→∞) n{1−(n^2 /(n^2 +1))}=lim_(n→∞) n{1−(1/(1+(1/n^2 )))}=0 lim_(n→∞) n{1−nv_n } lim_(n→∞) n{1−(n^2 /(n^2 +n))} lim_(n→∞) n{1−(1/(1+(1/n)))}=0 so lim_(n→∞) n{1−nu_n }=0

$${u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{2}}+...+\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}} \\ $$$$\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}>\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}>\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}} \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}>\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{2}}>\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}} \\ $$$$... \\ $$$$... \\ $$$$\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}>\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}}>\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}} \\ $$$$ \\ $$$$\:\:\:\:\:{T}_{{n}} \leftarrow\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{1}}>{u}_{{n}} >\frac{{n}}{{n}^{\mathrm{2}} +{n}}\rightarrow{v}_{{n}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\left\{\mathrm{1}−{nT}_{{n}} \right\} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\left\{\mathrm{1}−\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} +\mathrm{1}}\right\}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}\right\}=\mathrm{0} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\left\{\mathrm{1}−{nv}_{{n}} \right\} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\left\{\mathrm{1}−\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} +{n}}\right\} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\right\}=\mathrm{0} \\ $$$${so}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\left\{\mathrm{1}−{nu}_{{n}} \right\}=\mathrm{0} \\ $$

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