Menu Close

find-n-2-1-n-n-2-1-2-

Tinku Tara
Pin




Question Number 65673 by mathmax by abdo last updated on 01/Aug/19
find Σ_(n=2) ^∞ (((−1)^n )/((n^2 −1)^2 ))
find∑n=2∞(−1)n(n2−1)2
Commented by mathmax by abdo last updated on 02/Aug/19
let S =Σ_(n=2) ^∞ (((−1)^n )/((n^2 −1)^2 )) let decompose F(x) =(1/((x^2 −1)^2 )) F(x) =(1/((x−1)^2 (x+1)^2 )) =(a/(x−1))+(b/((x−1)^2 )) +(c/((x+1))) +(d/((x+1)^2 )) b =lim_(x→1) (x−1)^2 F(x) =(1/4) d =lim_(x→−1) (x+1)^2 F(x) =(1/4) ⇒ F(x) =(a/(x−1)) +(1/(4(x−1)^2 )) +(c/(x+1)) +(1/(4(x+1)^2 )) lim_(x→+∞) xF(x) =0 =a+c ⇒c=−a ⇒ F(x) =(a/(x−1)) +(1/(4(x−1)^2 ))−(a/(x+1)) +(1/(4(x+1)^2 )) F(0) =1 =−a +(1/4) −a +(1/4) =−2a+(1/2) ⇒2a =(1/2)−1 =−(1/2) ⇒ a =−(1/4) ⇒F(x) =((−1)/(4(x−1))) +(1/(4(x−1)^2 )) +(1/(4(x+1))) +(1/(4(x+1)^2 )) S =Σ_(n=2) ^∞ (−1)^n F(n) =−(1/4)Σ_(n=2) ^∞ (((−1)^n )/(n−1)) +(1/4)Σ_(n=2) ^∞ (((−1)^n )/((n−1)^2 )) +(1/4)Σ_(n=2) ^∞ (((−1)^n )/((n+1))) +(1/4)Σ_(n=2) ^∞ (((−1)^n )/((n+1)^2 )) Σ_(n=2) ^∞ (((−1)^n )/(n−1)) =Σ_(n=1) ^∞ (((−1)^(n+1) )/n) =−Σ_(n=1) ^∞ (((−1)^n )/n) =ln(2) Σ_(n=2) ^∞ (((−1)^n )/(n+1)) =Σ_(n=3) ^∞ (((−1)^(n−1) )/n) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −1+(1/2) =ln(2)−(1/2) Σ_(n=2) ^∞ (((−1)^n )/((n−1)^2 )) =Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 ) =−Σ_(n=1) ^∞ (((−1)^n )/n^2 ) but we have Σ_(n=1) ^∞ (((−1)^n )/n^x ) =(2^(1−x) −1)ξ(x) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(2^(1−2) −1)ξ(2) =−(1/2)(π^2 /6) =−(π^2 /(12)) Σ_(n=2) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=3) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=3) ^∞ (((−1)^n )/n^2 ) =−{Σ_(n=1) ^∞ (((−1)^n )/n^2 )+1−(1/4)} =(3/4)−(−(π^2 /(12)))=(3/4) +(π^2 /(12)) ⇒ S =−(1/4)ln(2) +(1/4)((π^2 /(12))) +(1/4)(ln2−(1/2)) +(1/4)((3/4)+(π^2 /(12))) =(π^2 /(48))−(1/8) +(3/(16)) +(π^2 /(48)) =(π^2 /(24)) +(1/(16))
letS=∑n=2∞(−1)n(n2−1)2letdecomposeF(x)=1(x2−1)2F(x)=1(x−1)2(x+1)2=ax−1+b(x−1)2+c(x+1)+d(x+1)2b=limx→1(x−1)2F(x)=14d=limx→−1(x+1)2F(x)=14⇒F(x)=ax−1+14(x−1)2+cx+1+14(x+1)2limx→+∞xF(x)=0=a+c⇒c=−a⇒F(x)=ax−1+14(x−1)2−ax+1+14(x+1)2F(0)=1=−a+14−a+14=−2a+12⇒2a=12−1=−12⇒a=−14⇒F(x)=−14(x−1)+14(x−1)2+14(x+1)+14(x+1)2S=∑n=2∞(−1)nF(n)=−14∑n=2∞(−1)nn−1+14∑n=2∞(−1)n(n−1)2+14∑n=2∞(−1)n(n+1)+14∑n=2∞(−1)n(n+1)2∑n=2∞(−1)nn−1=∑n=1∞(−1)n+1n=−∑n=1∞(−1)nn=ln(2)∑n=2∞(−1)nn+1=∑n=3∞(−1)n−1n=∑n=1∞(−1)n−1n−1+12=ln(2)−12∑n=2∞(−1)n(n−1)2=∑n=1∞(−1)n+1n2=−∑n=1∞(−1)nn2butwehave∑n=1∞(−1)nnx=(21−x−1)ξ(x)⇒∑n=1∞(−1)nn2=(21−2−1)ξ(2)=−12π26=−π212∑n=2∞(−1)n(n+1)2=∑n=3∞(−1)n−1n2=−∑n=3∞(−1)nn2=−{∑n=1∞(−1)nn2+1−14}=34−(−π212)=34+π212⇒S=−14ln(2)+14(π212)+14(ln2−12)+14(34+π212)=π248−18+316+π248=π224+116

Pin