Question Number 145602 by mnjuly1970 last updated on 06/Jul/21

Answered by ajfour last updated on 06/Jul/21
![(ii) S=Σ_(n=1) ^∞ (1/(n(2n+3))) ((3S)/2)=Σ_(n=1) ^∞ ((2n+3−2n)/(2n(2n+3))) =Σ_(n=1) ^∞ ((1/(2n))−(1/(2n+3))) ln (1−x)=−(x+(x^2 /2)+(x^3 /3)+..) Σ_(n=1) ^∞ (1/(2n))=−(1/2)lim_(x→1) ln (1−x) ln (1+x)−ln (1−x) =2(x+(x^3 /3)+(x^5 /5)+...) lim_(x→1) [(1/2){ln (1+x)−ln (1−x)} −x−(x^3 /3)] =Σ_(n=1) ^∞ (1/(2n+3)) ((3S)/2)=−(1/2)lim ln (1−x)− lim_(x→1) [(1/2){ln (1+x)−ln (1−x)} −x−(x^3 /3)] ((3S)/2)= (4/3)−((ln 2)/2) S=(8/9)−((ln 2)/3)](https://www.tinkutara.com/question/Q145604.png)
Commented by mnjuly1970 last updated on 06/Jul/21

Answered by Olaf_Thorendsen last updated on 06/Jul/21

Commented by mnjuly1970 last updated on 06/Jul/21

Answered by qaz last updated on 06/Jul/21

Answered by Kamel last updated on 06/Jul/21

Commented by mnjuly1970 last updated on 06/Jul/21

Answered by ArielVyny last updated on 06/Jul/21
![∅=∫_0 ^1 ln(Γ(2+x))dx =∫_0 ^1 ln((x+1)!)dx x+1=t→dt=dx ∅=∫_1 ^2 ln(t!)dt=∫_1 ^2 ln[((t/e))^t (√(2πt))] ∫_1 ^2 ln((t/e))^t +∫_1 ^2 ln((√(2πt)))dt ∫_1 ^2 ln(e^(tln((t/e))) )+∫_1 ^2 ln((√(2π)))dt+(1/2)∫_1 ^2 ln(t)dt ∫_1 ^2 tln((t/e))+ln((√(2π)))+(1/2)[xln(x)−x]_1 ^2 ∫_1 ^2 tln(t)−∫_1 ^2 tdt+ln((√(2π)))+(1/2)(2ln(2)+1) ∫_1 ^2 tln(t)dt−(3/2)+(1/2)ln(2π)+((2ln2)/2)+(1/2) ∫_1 ^2 tln(t)dt du=ln(t) v=t u=tln(t)−t dv=1 ∫_1 ^2 tln(t)=[t^2 ln(t)−t]−∫tlntdt+∫tdt 2∫_1 ^2 tln(t)dt=[t^2 ln(t)−t+(1/2)t^2 ]_1 ^2 =4ln(2)−2+(1/2)4+1−(1/2) ∫_1 ^2 tln(t)dt=2ln(2)−1+1+(1/2)−(1/4) =2ln(2)+(1/4) ∅=2ln(2)+(1/4)−(3/2)+((ln(2π))/2)+((2ln(2))/2)+(1/2) ∅=2ln(2)+(1/4)−1+((ln(8π))/2) ∅=−(3/4)+((2ln(4))/2)+((ln(8π))/2) ∅=−(3/4)+(1/2)ln(8^3 π)](https://www.tinkutara.com/question/Q145621.png)
Commented by mathmax by abdo last updated on 06/Jul/21
