let S_n = ∫_0 ^n ((x(−1)^([x]) )/((x+1 −[x])^3 ))dx 1) calculate S_n 2) find lim_(n→+∞) S


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Question Number 39660 by abdo mathsup 649 cc last updated on 09/Jul/18

$l e t S_{n} = \int_{0}^{n} \frac{x {(- 1)}^{[x]}}{{(x + 1 - [x])}^{3}} d x$ $1) c a l c u l a t e S_{n}$ $2) f i n d {l i m}_{n \to + \infty} S_{n}$
Commented by maxmathsup by imad last updated on 11/Jul/18
$1) S_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) ((x(−1)^k )/((x+1−k)^3 ))dx=Σ_(k=0) ^(n−1) (−1)^k ∫_k ^(k+1) (x/((x+1−k)^3 ))dx but ∫_k ^(k+1) (x/((x+1−k)^2 ))dx = ∫_k ^(k+1) ((x+1 −k +k−1)/((x+1−k)^2 ))dx =∫_k ^(k+1) (dx/(x+1−k)) +(k−1)[−(1/(x+1−k))]_k ^(k+1) =[ln∣x+1−k∣]_k ^(k+1) +(k−1) {1 −(1/2)}=ln(2) +(1/2)(k−1) S_n = Σ_(k=0) ^(n−1) (−1)^k {ln(2) +(1/2)(k−1)} =ln(2) Σ_(k=0) ^(n−1) (−1)^k +(1/2) Σ_(k=0) ^(n−1) (−1)^k (k−1) but we have Σ_(k=0) ^(n−1) (−1)^k =((1−(−1)^n )/2) Σ_(k=0) ^(n−1) (−1)^k (k−1)^ =−1 +Σ_(k=1) ^(n−1) (−1)^k (k−1) =−1 +Σ_(k=0) ^(n−2) (−1)^(k+1) k =−1 −Σ_(k=0) ^(n−2) k(−1)^k =−1−Σ_(k=1) ^(n−2) k(−1)^k Σ_(k=0) ^N x^k =((x^(N+1) −1)/(x−1)) ⇒Σ_(k=1) ^N k x^(k−1) =((Nx^(N+1) −(N+1)x^n +1)/((x−1)^2 )) ⇒ Σ_(k=1) ^N k x^k =(x/((x−1)^2 )){ Nx^(N+1) −(N+1)x^N +1} ⇒ Σ_(k=1) ^(n−2) k(−1)^k = ((−1)/4){(n−2)(−1)^(n−1) −(n−1)(−1)^(n−2) +1} ⇒ S_n =((ln(2))/2){1−(−1)^n } +(1/2){ −1 +(1/4){(n−2)(−1)^(n−1) −(n−1)(−1)^(n−2) +1}}$
$1) S_{n} = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{x {(- 1)}^{k}}{{(x + 1 - k)}^{3}} d x = \sum_{k = 0}^{n - 1} {(- 1)}^{k} \int_{k}^{k + 1} \frac{x}{{(x + 1 - k)}^{3}} d x b u t$ $\int_{k}^{k + 1} \frac{x}{{(x + 1 - k)}^{2}} d x = \int_{k}^{k + 1} \frac{x + 1 - k + k - 1}{{(x + 1 - k)}^{2}} d x$ $= \int_{k}^{k + 1} \frac{d x}{x + 1 - k} + (k - 1) {[- \frac{1}{x + 1 - k}]}_{k}^{k + 1}$ $= {[l n ∣ x + 1 - k ∣]}_{k}^{k + 1} + (k - 1) {1 - \frac{1}{2}} = l n (2) + \frac{1}{2} (k - 1)$ $S_{n} = \sum_{k = 0}^{n - 1} {(- 1)}^{k} {l n (2) + \frac{1}{2} (k - 1)}$ $= l n (2) \sum_{k = 0}^{n - 1} {(- 1)}^{k} + \frac{1}{2} \sum_{k = 0}^{n - 1} {(- 1)}^{k} (k - 1) b u t w e h a v e$ $\sum_{k = 0}^{n - 1} {(- 1)}^{k} = \frac{1 - {(- 1)}^{n}}{2}$ $\sum_{k = 0}^{n - 1} {(- 1)}^{k} {(k - 1)}^{} = - 1 + \sum_{k = 1}^{n - 1} {(- 1)}^{k} (k - 1)$ $= - 1 + \sum_{k = 0}^{n - 2} {(- 1)}^{k + 1} k = - 1 - \sum_{k = 0}^{n - 2} k {(- 1)}^{k} = - 1 - \sum_{k = 1}^{n - 2} k {(- 1)}^{k}$ $\sum_{k = 0}^{N} x^{k} = \frac{x^{N + 1} - 1}{x - 1} \Rightarrow \sum_{k = 1}^{N} k x^{k - 1} = \frac{{N x}^{N + 1} - (N + 1) x^{n} + 1}{{(x - 1)}^{2}} \Rightarrow$ $\sum_{k = 1}^{N} k x^{k} = \frac{x}{{(x - 1)}^{2}} {{N x}^{N + 1} - (N + 1) x^{N} + 1} \Rightarrow$ $\sum_{k = 1}^{n - 2} k {(- 1)}^{k} = \frac{- 1}{4} {(n - 2) {(- 1)}^{n - 1} - (n - 1) {(- 1)}^{n - 2} + 1} \Rightarrow$ $S_{n} = \frac{l n (2)}{2} {1 - {(- 1)}^{n}} + \frac{1}{2} {- 1 + \frac{1}{4} {(n - 2) {(- 1)}^{n - 1} - (n - 1) {(- 1)}^{n - 2} + 1}}$
Commented by maxmathsup by imad last updated on 11/Jul/18
$2) we have S_(2n) =−(1/2) +(1/8){−(2n−2)−(2n−1) +1)} =−(1/2) +(1/8){−4n +4} =(1/2) +(1/2){−n+1}=1−(n/2) S_(2n+1) =ln(2) +(1/2){ −1 +(1/4)((2n−1)−(2n)(−1) +1)} =ln(2) +(1/2){−1 +(1/4)(4n +2)} and its clear that (S_n ) is not convergent!$
$2) w e h a v e S_{2 n} = - \frac{1}{2} + \frac{1}{8} {- (2 n - 2) - (2 n - 1) + 1)}$ $= - \frac{1}{2} + \frac{1}{8} {- 4 n + 4} = \frac{1}{2} + \frac{1}{2} {- n + 1} = 1 - \frac{n}{2}$ $S_{2 n + 1} = l n (2) + \frac{1}{2} {- 1 + \frac{1}{4} ((2 n - 1) - (2 n) (- 1) + 1)}$ $= l n (2) + \frac{1}{2} {- 1 + \frac{1}{4} (4 n + 2)} a n d i t s c l e a r t h a t (S_{n}) i s n o t c o n v e r g e n t!$
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