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let-U-n-1-nx-n-1-x-x-3-dx-with-n-1-1-find-U-n-interms-of-n-2-find-lim-n-U-n-3-study-the-serie-n-1-U-n-

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Question Number 61232 by maxmathsup by imad last updated on 30/May/19
let U_n =∫_1 ^(+∞) (([nx]−[(n−1)x])/x^3 ) dx with n≥1 1) find U_n interms of n 2) find lim_(n→+∞) U_n 3) study the serie Σ_(n=1) ^∞ U_n
letUn=∫1+∞[nx]−[(n−1)x]x3dxwithn⩾11)findUnintermsofn2)findlimn→+∞Un3)studytheserie∑n=1∞Un
Commented by maxmathsup by imad last updated on 01/Jun/19
1) we have U_n =∫_1 ^(+∞) (([nx])/x^3 )dx −∫_1 ^(+∞) (([(n−1)x])/x^3 ) dx ∫_1 ^(+∞) (([nx])/x^3 ) dx =_(nx=t) n^3 ∫_n ^(+∞) (([t])/t^3 ) (dt/n) =n^2 ∫_n ^(+∞) (([t])/t^3 ) dt =n^2 Σ_(k=n) ^(+∞) ∫_k ^(k+1) (k/t^3 ) dt =n^2 Σ_(k=n) ^∞ k ∫_k ^(k+1) t^(−3) dt =n^2 Σ_(k=n) ^∞ k[−(1/2) t^(−2) ]_k ^(k+1) =−(n^2 /2) Σ_(k=n) ^∞ k{ (1/((k+1)^2 )) −(1/k^2 )} =−(n^2 /2){ Σ_(k=n) ^∞ (k/((k+1)^2 )) −Σ_(k=n) ^∞ (1/k)} we have Σ_(k=n) ^∞ (k/((k+1)^2 )) =Σ_(k=n+1) ^∞ ((k−1)/k^2 ) =Σ_(k=n+1) ^∞ (1/k) −Σ_(k=n+1) ^∞ (1/k^2 ) ⇒ ∫_1 ^(+∞) (([nx])/x^3 ) dx =−(n^2 /2){ Σ_(k=n) ^∞ (1/k) −(1/(n+1)) −Σ_(k=n+1) ^∞ (1/k^2 ) −Σ_(k=n) ^∞ (1/k)} =(n^2 /2) Σ_(k=n+1) ^∞ (1/k^2 ) + (n^2 /(2(n+1))) ∫_1 ^∞ (([(n−1)x])/x^3 ) dx =_((n−1)x =t) (n−1)^3 ∫_(n−1) ^(+∞) (([t])/t^3 ) (dt/(n−1)) =(n−1)^2 ∫_(n−1) ^(+∞) (([t])/t^3 ) dt =(n−1)^2 Σ_(k=n−1) ^∞ ∫_k ^(k+1) (k/t^3 ) dt =(n−1)^2 Σ_(k=n−1) ^∞ k ∫_k ^(k+1) (dt/t^(−3) ) =(n−1)^2 Σ_(k=n−1) ^∞ k[−(1/(2t^2 ))]_m ^(k+1) =−(((n−1)^2 )/2) Σ_(k=n−1) ^∞ k{ (1/((k+1)^2 )) −(1/k^2 )} =−(((n−1)^2 )/2){ Σ_(k=n−1) ^∞ (k/((k+1)^2 )) −Σ_(k=n−1) ^∞ (1/k)} Σ_(k=n−1) ^∞ (k/((k+1)^2 )) =Σ_(k=n) ^∞ ((k−1)/k^2 ) =Σ_(k=n) ^∞ (1/k) −Σ_(k=n) ^∞ (1/k^2 ) ⇒ ∫_1 ^∞ (([(n−1)x])/x^3 ) dx =−(((n−1)^2 )/2){ Σ_(k=n) ^∞ (1/k) −Σ_(k=n) ^∞ (1/k^2 ) −(1/(n−1)) −Σ_(k=n) ^∞ (1/k)} =−(((n−1)^2 )/2){−Σ_(k=n) ^∞ (1/k^2 ) −(1/(n−1)) } =(((n−1)^2 )/2) Σ_(k=n) ^∞ (1/k^2 ) +((n−1)/2) ⇒ U_n =(n^2 /2) Σ_(k=n+1) ^∞ (1/k^2 ) +(n^2 /(2(n+1))) −(((n−1)^2 )/2) Σ_(k=n) ^∞ (1/k^2 ) −((n−1)/2) .
1)wehaveUn=∫1+∞[nx]x3dx−∫1+∞[(n−1)x]x3dx∫1+∞[nx]x3dx=nx=tn3∫n+∞[t]t3dtn=n2∫n+∞[t]t3dt=n2∑k=n+∞∫kk+1kt3dt=n2∑k=n∞k∫kk+1t−3dt=n2∑k=n∞k[−12t−2]kk+1=−n22∑k=n∞k{1(k+1)2−1k2}=−n22{∑k=n∞k(k+1)2−∑k=n∞1k}wehave∑k=n∞k(k+1)2=∑k=n+1∞k−1k2=∑k=n+1∞1k−∑k=n+1∞1k2⇒∫1+∞[nx]x3dx=−n22{∑k=n∞1k−1n+1−∑k=n+1∞1k2−∑k=n∞1k}=n22∑k=n+1∞1k2+n22(n+1)∫1∞[(n−1)x]x3dx=(n−1)x=t(n−1)3∫n−1+∞[t]t3dtn−1=(n−1)2∫n−1+∞[t]t3dt=(n−1)2∑k=n−1∞∫kk+1kt3dt=(n−1)2∑k=n−1∞k∫kk+1dtt−3=(n−1)2∑k=n−1∞k[−12t2]mk+1=−(n−1)22∑k=n−1∞k{1(k+1)2−1k2}=−(n−1)22{∑k=n−1∞k(k+1)2−∑k=n−1∞1k}∑k=n−1∞k(k+1)2=∑k=n∞k−1k2=∑k=n∞1k−∑k=n∞1k2⇒∫1∞[(n−1)x]x3dx=−(n−1)22{∑k=n∞1k−∑k=n∞1k2−1n−1−∑k=n∞1k}=−(n−1)22{−∑k=n∞1k2−1n−1}=(n−1)22∑k=n∞1k2+n−12⇒Un=n22∑k=n+1∞1k2+n22(n+1)−(n−1)22∑k=n∞1k2−n−12.
Commented by maxmathsup by imad last updated on 02/Jun/19
we have ξ(2) =Σ_(k=1) ^∞ (1/k^2 ) =Σ_(k=1) ^n (1/k^2 ) +Σ_(k=n+1) ^∞ (1/k^2 ) =ξ_n (2) +Σ_(k=n+1) ^∞ (1/k^2 ) ⇒ Σ_(k=n+1) ^∞ (1/k^2 ) =ξ(2)−ξ_n (2) also Σ_(k=n) ^∞ (1/k^2 ) =ξ(2)−ξ_(n−1) (2) ⇒ U_n =(n^2 /2)( ξ(2)−ξ_n (2))−(((n−1)^2 )/2) (ξ(2)−ξ_(n−1) (2)) +(n^2 /(2(n+1))) −((n−1)/2) =((n^2 −(n−1)^2 )/2)ξ(2) −(n^2 /2)ξ_n (2)−(((n−1)^2 )/2) ξ_(n−1) (2) +((n^2 −(n−1)(n+1))/(2(n+1))) =((2n−1)/2) ξ(2) −(n^2 /2)( ξ_(n−1) (2) +(1/n^2 ))−(((n−1)^2 )/2) ξ_(n−1) (2) + (1/(2(n+1))) =(n−(1/2))ξ(2) −((n^2 +n^2 −2n+1)/2) ξ_(n−1) (2)−(n^2 /2) ξ_(n−1) (2) +(1/(2(n+1))) =(n−(1/2))ξ(2)−((2n^2 −2n +1)/2) ξ_(n−1) (2)−(n^2 /2) ξ_(n−1) (2)+(1/(2(n+1)))
wehaveξ(2)=∑k=1∞1k2=∑k=1n1k2+∑k=n+1∞1k2=ξn(2)+∑k=n+1∞1k2⇒∑k=n+1∞1k2=ξ(2)−ξn(2)also∑k=n∞1k2=ξ(2)−ξn−1(2)⇒Un=n22(ξ(2)−ξn(2))−(n−1)22(ξ(2)−ξn−1(2))+n22(n+1)−n−12=n2−(n−1)22ξ(2)−n22ξn(2)−(n−1)22ξn−1(2)+n2−(n−1)(n+1)2(n+1)=2n−12ξ(2)−n22(ξn−1(2)+1n2)−(n−1)22ξn−1(2)+12(n+1)=(n−12)ξ(2)−n2+n2−2n+12ξn−1(2)−n22ξn−1(2)+12(n+1)=(n−12)ξ(2)−2n2−2n+12ξn−1(2)−n22ξn−1(2)+12(n+1)
Commented by maxmathsup by imad last updated on 02/Jun/19
ξ_n (x) =Σ_(k=1) ^n (1/k^x ) with x>1
ξn(x)=∑k=1n1kxwithx>1

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