Menu Close

let-S-n-0-n-x-1-x-x-1-x-3-dx-1-calculate-S-n-2-find-lim-n-S-n-




Question Number 39660 by abdo mathsup 649 cc last updated on 09/Jul/18
let  S_n   = ∫_0 ^n      ((x(−1)^([x]) )/((x+1 −[x])^3 ))dx  1) calculate  S_n   2) find lim_(n→+∞)   S_n
$${let}\:\:{S}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{{n}} \:\:\:\:\:\frac{{x}\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\left({x}+\mathrm{1}\:−\left[{x}\right]\right)^{\mathrm{3}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{S}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:\:{S}_{{n}} \\ $$
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}}
$$\left.\mathrm{1}\right)\:{S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{{x}\left(−\mathrm{1}\right)^{{k}} }{\left({x}+\mathrm{1}−{k}\right)^{\mathrm{3}} }{dx}=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{k}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\frac{{x}}{\left({x}+\mathrm{1}−{k}\right)^{\mathrm{3}} }{dx}\:{but} \\ $$$$\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\:\frac{{x}}{\left({x}+\mathrm{1}−{k}\right)^{\mathrm{2}} }{dx}\:=\:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{x}+\mathrm{1}\:−{k}\:\:+{k}−\mathrm{1}}{\left({x}+\mathrm{1}−{k}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{dx}}{{x}+\mathrm{1}−{k}}\:\:+\left({k}−\mathrm{1}\right)\left[−\frac{\mathrm{1}}{{x}+\mathrm{1}−{k}}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\left[{ln}\mid{x}+\mathrm{1}−{k}\mid\right]_{{k}} ^{{k}+\mathrm{1}} \:\:+\left({k}−\mathrm{1}\right)\:\left\{\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2}}\right\}={ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left({k}−\mathrm{1}\right) \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \:\left\{{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left({k}−\mathrm{1}\right)\right\} \\ $$$$={ln}\left(\mathrm{2}\right)\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{k}} \:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \left({k}−\mathrm{1}\right)\:\:{but}\:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \:=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \left({k}−\mathrm{1}\right)^{} \:=−\mathrm{1}\:+\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \left({k}−\mathrm{1}\right) \\ $$$$=−\mathrm{1}\:+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{2}} \left(−\mathrm{1}\right)^{{k}+\mathrm{1}} {k}\:=−\mathrm{1}\:\:−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{2}} \:{k}\left(−\mathrm{1}\right)^{{k}} \:=−\mathrm{1}−\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{2}} {k}\left(−\mathrm{1}\right)^{{k}} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{N}} \:{x}^{{k}} \:=\frac{{x}^{{N}+\mathrm{1}} −\mathrm{1}}{{x}−\mathrm{1}}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{N}} \:{k}\:{x}^{{k}−\mathrm{1}} \:\:=\frac{{Nx}^{{N}+\mathrm{1}} −\left({N}+\mathrm{1}\right){x}^{{n}} \:+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{N}} \:{k}\:{x}^{{k}} \:\:=\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\left\{\:{Nx}^{{N}+\mathrm{1}} −\left({N}+\mathrm{1}\right){x}^{{N}} \:+\mathrm{1}\right\}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{2}} \:{k}\left(−\mathrm{1}\right)^{{k}} \:=\:\frac{−\mathrm{1}}{\mathrm{4}}\left\{\left({n}−\mathrm{2}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} −\left({n}−\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \:+\mathrm{1}\right\}\:\Rightarrow \\ $$$${S}_{{n}} =\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\left\{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right\}\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\:−\mathrm{1}\:\:+\frac{\mathrm{1}}{\mathrm{4}}\left\{\left({n}−\mathrm{2}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:−\left({n}−\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \:+\mathrm{1}\right\}\right\} \\ $$$$ \\ $$
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!
$$\left.\mathrm{2}\left.\right)\:{we}\:{have}\:\:{S}_{\mathrm{2}{n}} =−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{−\left(\mathrm{2}{n}−\mathrm{2}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)\:+\mathrm{1}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{−\mathrm{4}{n}\:+\mathrm{4}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{−{n}+\mathrm{1}\right\}=\mathrm{1}−\frac{{n}}{\mathrm{2}} \\ $$$${S}_{\mathrm{2}{n}+\mathrm{1}} \:\:={ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\:−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\left(\mathrm{2}{n}−\mathrm{1}\right)−\left(\mathrm{2}{n}\right)\left(−\mathrm{1}\right)\:+\mathrm{1}\right)\right\} \\ $$$$={ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{4}{n}\:+\mathrm{2}\right)\right\}\:{and}\:{its}\:{clear}\:{that}\:\left({S}_{{n}} \right)\:{is}\:{not}\:{convergent}! \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *