let A_n = ∫_0 ^n e^(−n( x+2−[x])) dx with n integr natural 1) calculate A_n 2) find lim_(n→+∞) A_n 3) study the convergence of Σ_n A


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Question Number 40040 by abdo mathsup 649 cc last updated on 15/Jul/18

$l e t A_{n} = \int_{0}^{n} e^{- n (x + 2 - [x])} d x w i t h n i n t e g r n a t u r a l$ $1) c a l c u l a t e A_{n}$ $2) f i n d {l i m}_{n \to + \infty} A_{n}$ $3) s t u d y t h e c o n v e r g e n c e o f \sum_{n} A_{n}$
Commented by abdo mathsup 649 cc last updated on 17/Jul/18
$1) we have A_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) e^(−n(x+2−k)) dx = Σ_(k=0) ^(n−1) e^(−n(2−k)) ∫_k ^(k+1) e^(−nx) dx =Σ_(k=0) ^(n−1) e^(n(k−2)) [−(1/n) e^(−nx) ]_k ^(k+1) = (1/n)Σ_(k=0) ^(n−1) e^(n(k−2)) { e^(−nk) −e^(−(n+1)k) } =(1/n) Σ_(k=0) ^(n−1) e^(−2n) −(1/n) Σ_(k=0) ^(n−1) e^(−2n−k) = e^(−2n) −(e^(−2n) /n) Σ_(k=0) ^(n−1) (e^(−1) )^k A_n =e^(−2n) −(e^(−2n) /n) ((1−(e^(−1) )^n )/(1−e^(−1) )) 2) we have lim_(n→+∞) e^(−2n) =0 lim_(n→+∞) (e^(−2n) /n) =0 and lim_(n→+∞) ((1− e^(−n) )/(2−e^(−1) )) =(1/(1−e^(−1) )) ⇒ lim_(n→+∞) A_n =0 3) we have Σ_(n=1) ^∞ A_n = Σ_(n=1) ^∞ e^(−2n) −(1/(1−e^(−1) ))Σ_(n=1) ^∞ (e^(−2n) /n) + (1/(1−e^(−1) )) Σ_(n=1) ^∞ (e^(−3n) /n) but we have Σ_(n=1) ^∞ e^(−2n) =Σ_(n=1) ^∞ (e^(−2) )^n =(1/(1−e^(−2) )) Σ_(n=1) ^∞ (e^(−2n) /n) =Σ_(n=1) ^∞ (((e^(−2) )^n )/n) =−ln(1−e^(−2) ) Σ_(n=1) ^∞ (e^(−3n) /n) =−ln(1−e^(−3) ) ⇒Σ A_n is convergent and Σ_(n≥1) A_n = (1/(1−e^(−2) )) +(1/(1−e^(−1) ))ln(1−e^(−2) ) −(1/(1−e^(−1) ))ln(1−e^(−3) ) .$
$1) w e h a v e A_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} e^{- n (x + 2 - k)} d x$ $= \sum_{k = 0}^{n - 1} e^{- n (2 - k)} \int_{k}^{k + 1} e^{- n x} d x$ $= \sum_{k = 0}^{n - 1} e^{n (k - 2)} {[- \frac{1}{n} e^{- n x}]}_{k}^{k + 1}$ $= \frac{1}{n} \sum_{k = 0}^{n - 1} e^{n (k - 2)} {e^{- n k} - e^{- (n + 1) k}}$ $= \frac{1}{n} \sum_{k = 0}^{n - 1} e^{- 2 n} - \frac{1}{n} \sum_{k = 0}^{n - 1} e^{- 2 n - k}$ $= e^{- 2 n} - \frac{e^{- 2 n}}{n} \sum_{k = 0}^{n - 1} {(e^{- 1})}^{k}$ $A_{n} = e^{- 2 n} - \frac{e^{- 2 n}}{n} \frac{1 - {(e^{- 1})}^{n}}{1 - e^{- 1}}$ $2) w e h a v e {l i m}_{n \to + \infty} e^{- 2 n} = 0$ ${l i m}_{n \to + \infty} \frac{e^{- 2 n}}{n} = 0 a n d {l i m}_{n \to + \infty} \frac{1 - e^{- n}}{2 - e^{- 1}} = \frac{1}{1 - e^{- 1}}$ $\Rightarrow {l i m}_{n \to + \infty} A_{n} = 0$ $3) w e h a v e \sum_{n = 1}^{\infty} A_{n} = \sum_{n = 1}^{\infty} e^{- 2 n} - \frac{1}{1 - e^{- 1}} \sum_{n = 1}^{\infty} \frac{e^{- 2 n}}{n}$ $+ \frac{1}{1 - e^{- 1}} \sum_{n = 1}^{\infty} \frac{e^{- 3 n}}{n} b u t w e h a v e$ $\sum_{n = 1}^{\infty} e^{- 2 n} = \sum_{n = 1}^{\infty} {(e^{- 2})}^{n} = \frac{1}{1 - e^{- 2}}$ $\sum_{n = 1}^{\infty} \frac{e^{- 2 n}}{n} = \sum_{n = 1}^{\infty} \frac{{(e^{- 2})}^{n}}{n} = - l n (1 - e^{- 2})$ $\sum_{n = 1}^{\infty} \frac{e^{- 3 n}}{n} = - l n (1 - e^{- 3}) \Rightarrow Σ A_{n} i s c o n v e r g e n t$ $a n d \sum_{n ⩾ 1} A_{n} = \frac{1}{1 - e^{- 2}} + \frac{1}{1 - e^{- 1}} l n (1 - e^{- 2})$ $- \frac{1}{1 - e^{- 1}} l n (1 - e^{- 3}) .$
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