let A_n = ∫_0 ^n ((x[x])/(1+x^2 )) dx 1) calculate A_n 2) find lim_(n→+∞) A


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Question Number 38896 by math khazana by abdo last updated on 01/Jul/18

$l e t A_{n} = \int_{0}^{n} \frac{x [x]}{1 + x^{2}} d x$ $1) c a l c u l a t e A_{n}$ $2) f i n d {l i m}_{n \to + \infty} A_{n}$
Commented by math khazana by abdo last updated on 01/Jul/18
$1) we have A_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) ((kx)/(1+x^2 ))dx =Σ_(k=0) ^(n−1) k ∫_k ^(k+1) ((xdx)/(1+x^2 )) =Σ_(k=0) ^(n−1) (k/2)(ln(1+(k+1)^2 ) −ln(1+k^2 )) =Σ_(k=0) ^(n−1) (k/2)ln{((1 +k^2 +2k+1)/(k^2 +1))} =Σ_(k=0) ^(n−1) (k/2)ln{ ((k^2 +2k+2)/(k^2 +1))} 2) ln{ ((k^2 +2k +2)/(k^2 +1))}=ln{ ((k^2 +1 +2k+1)/(k^2 +1))} =ln( 1+ ((2k+1)/(k^2 +1))) but ln^′ (1+x) =(1/(1+x)) =1−x +o(x^2 ) ⇒ ln(1+x) =x −(x^2 /2) +o(x^3 ) ⇒ x−(x^2 /2) ≤ ln(1+x)≤x ⇒ ((2k+1)/(k^2 +1)) −(1/2)(((2k+1)^2 )/((k^2 +1)^2 )) ≤ ln(1+((2k+1)/(k^2 +1)))≤ ((2k+1)/(k^2 +1)) ⇒ (k/2){ ((2k+1)/(k^2 +1)) −(1/2) (((2k+1)^2 )/((k^2 +1)^2 ))}≤ (k/2)ln(1+((2k+1)/(k^2 +1))) ≤ (k/2) ((2k+1)/(k^2 +1)) ⇒ A_n ≥ Σ_(k=1) ^(n−1) ((k(2k+1))/(2k^2 +2)) −Σ_(k=1) ^(n−1) (k/4) (((2k+1)^2 )/((k^2 +1)^2 )) →+∞ (n→+∞) so lim_(n→+∞) A_n =+∞.$
$1) w e h a v e A_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{k x}{1 + x^{2}} d x$ $= \sum_{k = 0}^{n - 1} k \int_{k}^{k + 1} \frac{x d x}{1 + x^{2}}$ $= \sum_{k = 0}^{n - 1} \frac{k}{2} (l n (1 + {(k + 1)}^{2}) - l n (1 + k^{2}))$ $= \sum_{k = 0}^{n - 1} \frac{k}{2} l n {\frac{1 + k^{2} + 2 k + 1}{k^{2} + 1}}$ $= \sum_{k = 0}^{n - 1} \frac{k}{2} l n {\frac{k^{2} + 2 k + 2}{k^{2} + 1}}$ $2) l n {\frac{k^{2} + 2 k + 2}{k^{2} + 1}} = l n {\frac{k^{2} + 1 + 2 k + 1}{k^{2} + 1}}$ $= l n (1 + \frac{2 k + 1}{k^{2} + 1}) b u t$ ${l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = 1 - x + o (x^{2}) \Rightarrow$ $l n (1 + x) = x - \frac{x^{2}}{2} + o (x^{3}) \Rightarrow$ $x - \frac{x^{2}}{2} ⩽ l n (1 + x) ⩽ x \Rightarrow$ $\frac{2 k + 1}{k^{2} + 1} - \frac{1}{2} \frac{{(2 k + 1)}^{2}}{{(k^{2} + 1)}^{2}} ⩽ l n (1 + \frac{2 k + 1}{k^{2} + 1}) ⩽ \frac{2 k + 1}{k^{2} + 1} \Rightarrow$ $\frac{k}{2} {\frac{2 k + 1}{k^{2} + 1} - \frac{1}{2} \frac{{(2 k + 1)}^{2}}{{(k^{2} + 1)}^{2}}} ⩽ \frac{k}{2} l n (1 + \frac{2 k + 1}{k^{2} + 1})$ $⩽ \frac{k}{2} \frac{2 k + 1}{k^{2} + 1} \Rightarrow$ $A_{n} ⩾ \sum_{k = 1}^{n - 1} \frac{k (2 k + 1)}{2 k^{2} + 2} - \sum_{k = 1}^{n - 1} \frac{k}{4} \frac{{(2 k + 1)}^{2}}{{(k^{2} + 1)}^{2}} \to + \infty$ $(n \to + \infty) s o {l i m}_{n \to + \infty} A_{n} = + \infty .$
Commented by math khazana by abdo last updated on 01/Jul/18
$2) another method we have 0≤x≤n ⇒ 0≤x^2 ≤n^2 ⇒ 1≤1+x^2 ≤ 1+n^2 ⇒ (1/(1+x^2 )) ≥ (1/(1+n^2 )) ⇒ A_n ≥ (1/(1+n^2 )) ∫_0 ^n x[x]dx but ∫_0 ^n x[x]dx =Σ_(k=0) ^(n−1) ∫_k ^(k+1) kx dx =Σ_(k=0) ^(n−1) k { (((k+1)^2 )/2) −(k^2 /2)} =Σ_(k=0) ^(n−1) k { ((k^2 +2k +1−k^2 )/2)} =Σ_(k=0) ^(n−1) k ((2k+1)/2) =Σ_(k=0) ^(n−1) k^2 +(1/2) Σ_(k=0) ^(n−1) k = (((n−1)(n−1+1)(2(n−1)+1))/6) +(1/2) (((n−1)n)/2) =((n(n−1)(2n−1))/6) +(1/4)(n^2 −n) ⇒ A_n ≥ (1/(n^2 +1)){ ((n(n−1)(2n−1))/6) +(1/4)(n^2 −n)}→+∞ (n→+∞)$
$2) a n o t h e r m e t h o d w e h a v e 0 ⩽ x ⩽ n \Rightarrow$ $0 ⩽ x^{2} ⩽ n^{2} \Rightarrow 1 ⩽ 1 + x^{2} ⩽ 1 + n^{2} \Rightarrow \frac{1}{1 + x^{2}} ⩾ \frac{1}{1 + n^{2}}$ $\Rightarrow A_{n} ⩾ \frac{1}{1 + n^{2}} \int_{0}^{n} x [x] d x b u t$ $\int_{0}^{n} x [x] d x = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} k x d x$ $= \sum_{k = 0}^{n - 1} k {\frac{{(k + 1)}^{2}}{2} - \frac{k^{2}}{2}}$ $= \sum_{k = 0}^{n - 1} k {\frac{k^{2} + 2 k + 1 - k^{2}}{2}}$ $= \sum_{k = 0}^{n - 1} k \frac{2 k + 1}{2}$ $= \sum_{k = 0}^{n - 1} k^{2} + \frac{1}{2} \sum_{k = 0}^{n - 1} k$ $= \frac{(n - 1) (n - 1 + 1) (2 (n - 1) + 1)}{6} + \frac{1}{2} \frac{(n - 1) n}{2}$ $= \frac{n (n - 1) (2 n - 1)}{6} + \frac{1}{4} (n^{2} - n) \Rightarrow$ $A_{n} ⩾ \frac{1}{n^{2} + 1} {\frac{n (n - 1) (2 n - 1)}{6} + \frac{1}{4} (n^{2} - n)} \to + \infty$ $(n \to + \infty)$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18
	$A_n =∫_0 ^1 ((x×0)/(1+x^2 ))dx+∫_1 ^2 ((x×1)/(1+x^2 ))dx+∫_2 ^3 ((x×2)/(1+x^2 ))+ ∫_(n−1) ^n ((x×(n−1))/(1+x^2 ))dx =(1/2)×1∫_1 ^2 ((2x)/(1+x^2 ))dx+(1/2)×2∫_2 ^3 ((2x)/(1+x^2 ))dx+(1/2)×3∫_3 ^4 ((2x)/(1+x^2 )) (1/2)×4∫_4 ^5 ((2x)/(1+x^2 ))+...+(1/2)×(n−1)∫_(n−1) ^n ((2x)/(1+x^2 ))dx =(1/2){1×∣ln(1+x^2 )∣_1 ^2 +2×∣ln(1+x^2 )∣_2 ^3 +..+ (n−1)∣ln(1+x^2 )∣_(n−1) ^n } =(1/2)[1×(ln5−ln2)+2×(ln10−ln5)+ +...+(n−1){ln(1+n^2 )−ln(1+(n−1)^2 }] =(1/2)ln{((5/2))^1 ×(((10)/5))^2 ×...×(((1+n^2 )/(1+(n−1)^2 )))^(n−1) }$
	$A_{n} = \int_{0}^{1} \frac{x \times 0}{1 + x^{2}} d x + \int_{1}^{2} \frac{x \times 1}{1 + x^{2}} d x + \int_{2}^{3} \frac{x \times 2}{1 + x^{2}} +$ $\int_{n - 1}^{n} \frac{x \times (n - 1)}{1 + x^{2}} d x$ $= \frac{1}{2} \times 1 \int_{1}^{2} \frac{2 x}{1 + x^{2}} d x + \frac{1}{2} \times 2 \int_{2}^{3} \frac{2 x}{1 + x^{2}} d x + \frac{1}{2} \times 3 \int_{3}^{4} \frac{2 x}{1 + x^{2}}$ $\frac{1}{2} \times 4 \int_{4}^{5} \frac{2 x}{1 + x^{2}} + . . . + \frac{1}{2} \times (n - 1) \int_{n - 1}^{n} \frac{2 x}{1 + x^{2}} d x$ $= \frac{1}{2} {1 \times ∣ l n (1 + x^{2}) ∣_{1}^{2} + 2 \times ∣ l n (1 + x^{2}) ∣_{2}^{3} + . . +$ $(n - 1) ∣ l n (1 + x^{2}) ∣_{n - 1}^{n}}$ $= \frac{1}{2} [1 \times (l n 5 - l n 2) + 2 \times (l n 10 - l n 5) +$ $+ . . . + (n - 1) {l n (1 + n^{2}) - l n (1 + {(n - 1)}^{2}}]$ $= \frac{1}{2} l n {{(\frac{5}{2})}^{1} \times {(\frac{10}{5})}^{2} \times . . . \times {(\frac{1 + n^{2}}{1 + {(n - 1)}^{2}})}^{n - 1}}$
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