Menu Close

calculate-n-1-1-n-n-2-n-1-3-

Tinku Tara 0 Comments
Pin




Question Number 69376 by mathmax by abdo last updated on 22/Sep/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^2 (n+1)^3 ))
calculate∑n=1∞(−1)nn2(n+1)3
Commented by mathmax by abdo last updated on 24/Sep/19
let S_n =Σ_(k=1) ^n (((−1)^k )/(k^2 (k+1)^3 )) first let decompose F(x)=(1/(x^2 (x+1)^3 )) F(x)=(a/x) +(b/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(e/((x+1)^3 )) b=lim_(x→0) x^2 F(x)=1 e=lim_(x→−1) (x+1)^3 F(x)=1 ⇒F(x)=(a/x)+(1/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x)=(a/x)−(a/(x+1)) +(1/x^2 ) +(d/((x+1)^2 )) +(1/((x+1)^3 )) =(a/(x^2 +x)) +(1/x^2 ) +(d/((x+1)^2 )) +(1/((x+1)^3 )) ⇒lim_(x→+∞) x^2 F(x) =a+1 +d=0 ⇒d =−a−1 ⇒ F(x)=(a/x)−(a/(x+1)) +(1/x^2 )−((a+1)/((x+1)^2 )) +(1/((x+1)^3 )) F(−2) =−(1/4) =−(a/2) +a +(1/4)−a−1−1 =−(a/2)−(7/4) ⇒ (1/4) =(a/2)+(7/4) ⇒1 =2a+7 ⇒2a=−6 ⇒a=−3 ⇒ F(x)=−(3/x) +(3/(x+1)) +(1/x^2 ) +(2/((x+1)^2 )) +(1/((x+1)^3 )) ⇒ S_n =Σ_(k=1) ^n (−1)^k F(k)=−3Σ_(k=1) ^n (((−1)^k )/k) +3Σ_(k=1) ^n (((−1)^k )/(k+1)) +Σ_(k=1) ^n (((−1)^k )/k^2 ) +2Σ_(k=1) ^n (((−1)^k )/((k+1)^2 )) +Σ_(k=1) ^n (((−1)^k )/((k+1)^3 )) Σ_(k=1) ^n (((−1)^k )/k) →−ln(2) Σ_(k=1) ^n (((−1)^k )/(k+1)) =Σ_(k=2) ^(n+1) (((−1)^(k−1) )/k) →ln(2)−1 Σ_(k=1) ^n (((−1)^k )/k^2 ) →Σ_(k=1) ^∞ (((−1)^k )/k^2 ) =(2^(1−2) −1)ξ(2)=−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(k=1) ^n (((−1)^k )/((k+1)^2 )) =Σ_(k=2) ^(n+1) (((−1)^(k−1) )/k^2 ) =−Σ_(k=1) ^(n+1) (((−1)^k )/k^2 )−1→(π^2 /(12))−1 Σ_(k=1) ^n (((−1)^k )/((k+1)^3 )) =Σ_(k=2) ^(n+1) (((−1)^(k−1) )/k^3 ) =−Σ_(k=1) ^(n+1) (((−1)^k )/k^3 ) −1 →−(2^(1−3) −1)ξ(3)−1 =(1−(1/4))ξ(3)−1 =(3/4)ξ(3)−1 ⇒ S =lim_(n→+∞) S_n =3ln(2)+3ln(2)−3−(π^2 /(12)) +(π^2 /6)−2 +(3/4)ξ(3)−1 S=6ln(2) +(π^2 /(12)) +(3/4)ξ(3) −6
letSn=∑k=1n(−1)kk2(k+1)3firstletdecomposeF(x)=1x2(x+1)3F(x)=ax+bx2+cx+1+d(x+1)2+e(x+1)3b=limx→0x2F(x)=1e=limx→−1(x+1)3F(x)=1⇒F(x)=ax+1x2+cx+1+d(x+1)2+1(x+1)3limx→+∞xF(x)=0=a+c⇒c=−a⇒F(x)=ax−ax+1+1x2+d(x+1)2+1(x+1)3=ax2+x+1x2+d(x+1)2+1(x+1)3⇒limx→+∞x2F(x)=a+1+d=0⇒d=−a−1⇒F(x)=ax−ax+1+1x2−a+1(x+1)2+1(x+1)3F(−2)=−14=−a2+a+14−a−1−1=−a2−74⇒14=a2+74⇒1=2a+7⇒2a=−6⇒a=−3⇒F(x)=−3x+3x+1+1x2+2(x+1)2+1(x+1)3⇒Sn=∑k=1n(−1)kF(k)=−3∑k=1n(−1)kk+3∑k=1n(−1)kk+1+∑k=1n(−1)kk2+2∑k=1n(−1)k(k+1)2+∑k=1n(−1)k(k+1)3∑k=1n(−1)kk→−ln(2)∑k=1n(−1)kk+1=∑k=2n+1(−1)k−1k→ln(2)−1∑k=1n(−1)kk2→∑k=1∞(−1)kk2=(21−2−1)ξ(2)=−12π26=−π212∑k=1n(−1)k(k+1)2=∑k=2n+1(−1)k−1k2=−∑k=1n+1(−1)kk2−1→π212−1∑k=1n(−1)k(k+1)3=∑k=2n+1(−1)k−1k3=−∑k=1n+1(−1)kk3−1→−(21−3−1)ξ(3)−1=(1−14)ξ(3)−1=34ξ(3)−1⇒S=limn→+∞Sn=3ln(2)+3ln(2)−3−π212+π26−2+34ξ(3)−1S=6ln(2)+π212+34ξ(3)−6

Pin

Leave a Reply

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