let u_n =∫_0 ^∞ ((t−[t])/(t(t+n)))dt find a equivalent of u


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Question Number 44317 by abdo.msup.com last updated on 26/Sep/18

$l e t u_{n} = \int_{0}^{\infty} \frac{t - [t]}{t (t + n)} d t$ $f i n d a e q u i v a l e n t o f u_{n} w h e n n \to + \infty$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Sep/18
	$((t−r)/(t(t+n)))=(1/(t+n))−(r/(t(t+n))) =(1/(t+n))−(r/n)×(((t+n)−t)/(t(t+n))) =(1/(t+n))−(r/n)×((1/t)−(1/(t+n))) =(1/(t+n))+(r/n)×(1/(t+n))−(r/n)×(1/t) =(1+(r/n))((1/(t+n)))−(r/n)×(1/t) ∫_0 ^1 {(1+(0/n))((1/(t+n)))−(0/n)×(1/t)}dt+ ∫_1 ^2 (1+(1/n))((1/(t+n)))−(1/n)×(1/t)dt+... ∫_r ^(r+1) (1+(r/n))((1/(t+n)))−(r/n)×(1/t)+... =(1+(0/n))∣ln(t+n)∣_0 ^1 −(0/n)×∣lnt∣_0 ^1 + (1+(1/n))∣ln(t+n)∣_1 ^2 −(1/n)×∣lnt∣_1 ^2 +... (1+(r/n))∣ln(t+n)∣_r ^(r+1) +(r/n)×∣lnt∣_r ^(r+1) +... =(1+(0/n)){ln(((1+n)/(0+n)))}−(0/n)×{ln1−ln0}+ (1+(1/n)){ln(((2+n)/(1+n)))}−(1/n)×{ln2−ln1}+.. (1+(r/n)){ln(((r+1+n)/(r+n)))}−(r/n){ln(r+1)−lnr}+.. =Σ_(r=0) ^∞ (1+(r/n)){ln(((r+1+n)/(r+n)))}−(r/n){ln(((r+1)/r))}$
	$\frac{t - r}{t (t + n)} = \frac{1}{t + n} - \frac{r}{t (t + n)}$ $= \frac{1}{t + n} - \frac{r}{n} \times \frac{(t + n) - t}{t (t + n)}$ $= \frac{1}{t + n} - \frac{r}{n} \times (\frac{1}{t} - \frac{1}{t + n})$ $= \frac{1}{t + n} + \frac{r}{n} \times \frac{1}{t + n} - \frac{r}{n} \times \frac{1}{t}$ $= (1 + \frac{r}{n}) (\frac{1}{t + n}) - \frac{r}{n} \times \frac{1}{t}$ $\int_{0}^{1} {(1 + \frac{0}{n}) (\frac{1}{t + n}) - \frac{0}{n} \times \frac{1}{t}} d t +$ $\int_{1}^{2} (1 + \frac{1}{n}) (\frac{1}{t + n}) - \frac{1}{n} \times \frac{1}{t} d t + . . .$ $\int_{r}^{r + 1} (1 + \frac{r}{n}) (\frac{1}{t + n}) - \frac{r}{n} \times \frac{1}{t} + . . .$ $= (1 + \frac{0}{n}) ∣ l n (t + n) ∣_{0}^{1} - \frac{0}{n} \times ∣ l n t ∣_{0}^{1} +$ $(1 + \frac{1}{n}) ∣ l n (t + n) ∣_{1}^{2} - \frac{1}{n} \times ∣ l n t ∣_{1}^{2} + . . .$ $(1 + \frac{r}{n}) ∣ l n (t + n) ∣_{r}^{r + 1} + \frac{r}{n} \times ∣ l n t ∣_{r}^{r + 1} + . . .$ $= (1 + \frac{0}{n}) {l n (\frac{1 + n}{0 + n})} - \frac{0}{n} \times {l n 1 - l n 0} +$ $(1 + \frac{1}{n}) {l n (\frac{2 + n}{1 + n})} - \frac{1}{n} \times {l n 2 - l n 1} + . .$ $(1 + \frac{r}{n}) {l n (\frac{r + 1 + n}{r + n})} - \frac{r}{n} {l n (r + 1) - l n r} + . .$ $= \underset{r = 0}{\sum^{\infty}} (1 + \frac{r}{n}) {l n (\frac{r + 1 + n}{r + n})} - \frac{r}{n} {l n (\frac{r + 1}{r})}$
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