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Question Number 44317 by abdo.msup.com last updated on 26/Sep/18
let u_n =∫_0 ^∞ ((t−[t])/(t(t+n)))dt find a equivalent of u_n when n→+∞
$${let}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}−\left[{t}\right]}{{t}\left({t}+{n}\right)}{dt} \\ $$$${find}\:{a}\:{equivalent}\:{of}\:{u}_{{n}} \:{when}\:{n}\rightarrow+\infty \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 26/Sep/18
((t−r)/(t(t+n)))=(1/(t+n))−(r/(t(t+n))) =(1/(t+n))−(r/n)×(((t+n)−t)/(t(t+n))) =(1/(t+n))−(r/n)×((1/t)−(1/(t+n))) =(1/(t+n))+(r/n)×(1/(t+n))−(r/n)×(1/t) =(1+(r/n))((1/(t+n)))−(r/n)×(1/t) ∫_0 ^1 {(1+(0/n))((1/(t+n)))−(0/n)×(1/t)}dt+ ∫_1 ^2 (1+(1/n))((1/(t+n)))−(1/n)×(1/t)dt+... ∫_r ^(r+1) (1+(r/n))((1/(t+n)))−(r/n)×(1/t)+... =(1+(0/n))∣ln(t+n)∣_0 ^1 −(0/n)×∣lnt∣_0 ^1 + (1+(1/n))∣ln(t+n)∣_1 ^2 −(1/n)×∣lnt∣_1 ^2 +... (1+(r/n))∣ln(t+n)∣_r ^(r+1) +(r/n)×∣lnt∣_r ^(r+1) +... =(1+(0/n)){ln(((1+n)/(0+n)))}−(0/n)×{ln1−ln0}+ (1+(1/n)){ln(((2+n)/(1+n)))}−(1/n)×{ln2−ln1}+.. (1+(r/n)){ln(((r+1+n)/(r+n)))}−(r/n){ln(r+1)−lnr}+.. =Σ_(r=0) ^∞ (1+(r/n)){ln(((r+1+n)/(r+n)))}−(r/n){ln(((r+1)/r))}
$$\frac{{t}−{r}}{{t}\left({t}+{n}\right)}=\frac{\mathrm{1}}{{t}+{n}}−\frac{{r}}{{t}\left({t}+{n}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{t}+{n}}−\frac{{r}}{{n}}×\frac{\left({t}+{n}\right)−{t}}{{t}\left({t}+{n}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{t}+{n}}−\frac{{r}}{{n}}×\left(\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}+{n}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{t}+{n}}+\frac{{r}}{{n}}×\frac{\mathrm{1}}{{t}+{n}}−\frac{{r}}{{n}}×\frac{\mathrm{1}}{{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}+\frac{{r}}{{n}}\right)\left(\frac{\mathrm{1}}{{t}+{n}}\right)−\frac{{r}}{{n}}×\frac{\mathrm{1}}{{t}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\left(\mathrm{1}+\frac{\mathrm{0}}{{n}}\right)\left(\frac{\mathrm{1}}{{t}+{n}}\right)−\frac{\mathrm{0}}{{n}}×\frac{\mathrm{1}}{{t}}\right\}{dt}+ \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\frac{\mathrm{1}}{{t}+{n}}\right)−\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{{t}}{dt}+… \\ $$$$\int_{{r}} ^{{r}+\mathrm{1}} \left(\mathrm{1}+\frac{{r}}{{n}}\right)\left(\frac{\mathrm{1}}{{t}+{n}}\right)−\frac{{r}}{{n}}×\frac{\mathrm{1}}{{t}}+… \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{0}}{{n}}\right)\mid{ln}\left({t}+{n}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{0}}{{n}}×\mid{lnt}\mid_{\mathrm{0}} ^{\mathrm{1}} + \\ $$$$\:\:\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\mid{ln}\left({t}+{n}\right)\mid_{\mathrm{1}} ^{\mathrm{2}} −\frac{\mathrm{1}}{{n}}×\mid{lnt}\mid_{\mathrm{1}} ^{\mathrm{2}} +… \\ $$$$\:\:\:\:\:\:\left(\mathrm{1}+\frac{{r}}{{n}}\right)\mid{ln}\left({t}+{n}\right)\mid_{{r}} ^{{r}+\mathrm{1}} +\frac{{r}}{{n}}×\mid{lnt}\mid_{{r}} ^{{r}+\mathrm{1}} +… \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{0}}{{n}}\right)\left\{{ln}\left(\frac{\mathrm{1}+{n}}{\mathrm{0}+{n}}\right)\right\}−\frac{\mathrm{0}}{{n}}×\left\{{ln}\mathrm{1}−{ln}\mathrm{0}\right\}+ \\ $$$$\:\:\:\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left\{{ln}\left(\frac{\mathrm{2}+{n}}{\mathrm{1}+{n}}\right)\right\}−\frac{\mathrm{1}}{{n}}×\left\{{ln}\mathrm{2}−{ln}\mathrm{1}\right\}+.. \\ $$$$\:\:\:\:\:\left(\mathrm{1}+\frac{{r}}{{n}}\right)\left\{{ln}\left(\frac{{r}+\mathrm{1}+{n}}{{r}+{n}}\right)\right\}−\frac{{r}}{{n}}\left\{{ln}\left({r}+\mathrm{1}\right)−{lnr}\right\}+.. \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\frac{{r}}{{n}}\right)\left\{{ln}\left(\frac{{r}+\mathrm{1}+{n}}{{r}+{n}}\right)\right\}−\frac{{r}}{{n}}\left\{{ln}\left(\frac{{r}+\mathrm{1}}{{r}}\right)\right\} \\ $$$$ \\ $$$$ \\ $$

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