let A_n =∫_0 ^n ((t[t])/(3+t^2 ))dt 1)calculate lim_(n→+∞) A_n 2) find nature if Σ A


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Question Number 57488 by Abdo msup. last updated on 05/Apr/19

$l e t A_{n} = \int_{0}^{n} \frac{t [t]}{3 + t^{2}} d t$ $1) c a l c u l a t e {l i m}_{n \to + \infty} A_{n}$ $2) f i n d n a t u r e i f Σ A_{n}$
Commented by maxmathsup by imad last updated on 06/Apr/19
$1) we have A_n =Σ_(k=0) ^(n−1) ∫_k ^(k+1) ((tk)/(3+t^2 )) dt =Σ_(k=0) ^(n−1) k ∫_k ^(k+1) ((tdt)/(t^2 +3)) =(1/2)Σ_(k=0) ^(n−1) k [ ln(t^2 +3)]_k ^(k+1) =(1/2) Σ_(k=0) ^(n−1) k ln{(((k+1)^2 +3)/(k^2 +3))} =(1/2) Σ_(k=0) ^(n−1) k ln{((k^2 +3 +2k)/(k^2 +3))} =(1/2) Σ_(k=1) ^(n−1) k ln{ 1+((2k)/(k^2 +3))} . =(1/2){ln(1+(2/4)) +2 ln( 1+(4/7)) +3 ln(1+(6/(12)))+....+(n−1)ln(1+((2n−2)/((n−1)^2 +3)))}$
$1) w e h a v e A_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{t k}{3 + t^{2}} d t = \sum_{k = 0}^{n - 1} k \int_{k}^{k + 1} \frac{t d t}{t^{2} + 3}$ $= \frac{1}{2} \sum_{k = 0}^{n - 1} k {[l n (t^{2} + 3)]}_{k}^{k + 1} = \frac{1}{2} \sum_{k = 0}^{n - 1} k l n {\frac{{(k + 1)}^{2} + 3}{k^{2} + 3}}$ $= \frac{1}{2} \sum_{k = 0}^{n - 1} k l n {\frac{k^{2} + 3 + 2 k}{k^{2} + 3}} = \frac{1}{2} \sum_{k = 1}^{n - 1} k l n {1 + \frac{2 k}{k^{2} + 3}} .$ $= \frac{1}{2} {l n (1 + \frac{2}{4}) + 2 l n (1 + \frac{4}{7}) + 3 l n (1 + \frac{6}{12}) + . . . . + (n - 1) l n (1 + \frac{2 n - 2}{{(n - 1)}^{2} + 3})}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Apr/19
	$∫_0 ^n ((t[t])/(3+t^2 ))dt (1/2)∫_0 ^1 ((d(3+t^2 ))/(3+t^2 ))×0+(1/2)∫_1 ^2 ((d(3+t^2 ))/(3+t^2 ))×1+(1/2)∫_2 ^3 ((d(3+t^2 ))/(3+t^2 ))×2dt+...+(1/2)∫_(n−1) ^n ((d(3+t^2 ))/(3+t^2 ))×(n−1)dt =(1/2)[0×∣ln(3+t^2 )∣_0 ^1 +1×∣ln(3+t^2 )∣_1 ^2 +2×∣ln(3+t^2 )∣_2 ^3 +..+(n−1)∣ln(3+t^2 )∣_(n−1) ^n ] =(1/2)Σ_(n=1) ^n (n−1)ln{((3+n^2 )/(3+(n−1)^2 ))}$
	$\int_{0}^{n} \frac{t [t]}{3 + t^{2}} d t$ $\frac{1}{2} \int_{0}^{1} \frac{d (3 + t^{2})}{3 + t^{2}} \times 0 + \frac{1}{2} \int_{1}^{2} \frac{d (3 + t^{2})}{3 + t^{2}} \times 1 + \frac{1}{2} \int_{2}^{3} \frac{d (3 + t^{2})}{3 + t^{2}} \times 2 d t + . . . + \frac{1}{2} \int_{n - 1}^{n} \frac{d (3 + t^{2})}{3 + t^{2}} \times (n - 1) d t$ $= \frac{1}{2} [0 \times ∣ l n (3 + t^{2}) ∣_{0}^{1} + 1 \times ∣ l n (3 + t^{2}) ∣_{1}^{2} + 2 \times ∣ l n (3 + t^{2}) ∣_{2}^{3} + . . + (n - 1) ∣ l n (3 + t^{2}) ∣_{n - 1}^{n}]$ $= \frac{1}{2} \underset{n = 1}{\sum^{n}} (n - 1) l n {\frac{3 + n^{2}}{3 + {(n - 1)}^{2}}}$
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