let-u-n-0-t-t-t-t-n-dt-find-a-equivalent-of-u-n-when-n-

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

\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 + \dots

\int_{r}^{r + 1} (1 + \frac{r}{n}) (\frac{1}{t + n}) - \frac{r}{n} \times \frac{1}{t} + \dots

= (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} + \dots

(1 + \frac{r}{n}) ∣ l n (t + n) ∣_{r}^{r + 1} + \frac{r}{n} \times ∣ l n t ∣_{r}^{r + 1} + \dots

= (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})}