Menu Close

find-the-value-of-n-2-3n-2-1-n-1-3-n-1-3-

Tinku Tara 0 Comments
Pin




Question Number 57489 by Abdo msup. last updated on 05/Apr/19
find the value of Σ_(n=2) ^∞ ((3n^2 +1)/((n−1)^3 (n+1)^3 ))
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by maxmathsup by imad last updated on 06/Apr/19
we have (n+1)^3 −(n−1)^3 =n^3 +3n^2 +3n+1 −(n^3 −3n^2 +3n−1) =6n^2 +2 =2(3n^2 +1) ⇒3n^2 +1 =(1/2){(n+1)^3 −(n−1)^3 } ⇒ S =(1/2)Σ_(n=2) ^∞ (((n+1)^3 −(n−1)^3 )/((n−1)^3 (n+1)^3 )) =(1/2) Σ_(n=2) ^∞ (1/((n−1)^3 )) −(1/2) Σ_(n=2) ^∞ (1/((n+1)^3 )) but Σ_(n=2) ^∞ (1/((n−1)^3 )) =Σ_(n=1) ^∞ (1/n^3 ) =ξ(3) Σ_(n=2) ^∞ (1/((n+1)^3 )) =Σ_(n=3) ^∞ (1/n^3 ) =Σ_(n=1) ^∞ (1/n^3 ) −1−(1/8) =ξ(3)−(9/8) ⇒ S =(1/2)ξ(3)−(1/2)(ξ(3)−(9/8)) =(9/(16)) ★ S =(9/(16)) ★
$${we}\:{have}\:\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} \:={n}^{\mathrm{3}} \:+\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{1}\:−\left({n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{3}{n}−\mathrm{1}\right) \\ $$$$=\mathrm{6}{n}^{\mathrm{2}} \:+\mathrm{2}\:=\mathrm{2}\left(\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{1}\right)\:\Rightarrow\mathrm{3}{n}^{\mathrm{2}} \:+\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} \right\}\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\left({n}−\mathrm{1}\right)^{\mathrm{3}} }{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${but}\:\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\xi\left(\mathrm{3}\right) \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{3}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\:=\xi\left(\mathrm{3}\right)−\frac{\mathrm{9}}{\mathrm{8}}\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{2}}\xi\left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\xi\left(\mathrm{3}\right)−\frac{\mathrm{9}}{\mathrm{8}}\right)\:=\frac{\mathrm{9}}{\mathrm{16}} \\ $$$$\bigstar\:{S}\:=\frac{\mathrm{9}}{\mathrm{16}}\:\bigstar \\ $$
Answered by Smail last updated on 06/Apr/19
F(n)((3n^2 +1)/((n−1)^3 (n+1)^3 ))=(a/(n−1))+(b/((n−1)^2 ))+(c/((n−1)^3 ))+(d/(n+1))+(e/((n+1)^2 ))+(f/((n+1)^3 )) F(−n)=−(a/(n+1))+(b/((n+1)^2 ))−(c/((n+1)^3 ))−(d/((n−1)))+(e/((n−1)^2 ))−(f/((n−1)^3 )) F(−n)=F(n) b=e ; a=−d ; c=−f=(1/2) F(0)=−1=−a+b−c+d+e+f b−a−c−a+b−c=−1⇔b−a−c=((−1)/2)⇔b=a F(2)=((13)/(27))=a+a+c−(a/3)+(a/9)−(c/(27)) =a(2−(2/9))+(1/2)(1−(1/(27)))=((13)/(27)) ((16)/9)a+((26)/(2×27))=((13)/(27))⇔a=0 a=b=e=d=0 ((3n^2 +1)/((n−1)^3 (n+1)^3 ))=(1/2)((1/((n−1)^3 ))−(1/((n+1)^3 ))) Σ_(n=2) ^∞ ((3n^2 +1)/((n−1)^3 (n+1)^3 ))=(1/2)Σ_(n=2) ^∞ ((1/((n−1)^3 ))−(1/((n+1)^3 ))) let m+1=n−1 Σ_(n=2) ^∞ ((3n^2 +1)/((n−1)^3 (n+1)^3 ))=(1/2)Σ_(m=0) ^∞ (1/((m+1)^3 ))−(1/2)Σ_(n=2) ^∞ (1/((n+1)^3 )) =(1/2)(Σ_(n=2) ^∞ (1/((n+1)^3 ))+1+(1/2^3 ))−(1/2)Σ_(n=2) ^∞ (1/((n+1)^3 )) =(1/2)(((2^3 +1)/2^3 )) Σ_(n=2) ^∞ ((3n^2 +1)/((n−1)^3 (n+1)^3 ))=(9/(16))
$${F}\left({n}\right)\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{{a}}{{n}−\mathrm{1}}+\frac{{b}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{c}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{{d}}{{n}+\mathrm{1}}+\frac{{e}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{f}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${F}\left(−{n}\right)=−\frac{{a}}{{n}+\mathrm{1}}+\frac{{b}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{{c}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{{d}}{\left({n}−\mathrm{1}\right)}+\frac{{e}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{{f}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${F}\left(−{n}\right)={F}\left({n}\right) \\ $$$${b}={e}\:\:;\:\:{a}=−{d}\:\:;\:{c}=−{f}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{0}\right)=−\mathrm{1}=−{a}+{b}−{c}+{d}+{e}+{f} \\ $$$${b}−{a}−{c}−{a}+{b}−{c}=−\mathrm{1}\Leftrightarrow{b}−{a}−{c}=\frac{−\mathrm{1}}{\mathrm{2}}\Leftrightarrow{b}={a} \\ $$$${F}\left(\mathrm{2}\right)=\frac{\mathrm{13}}{\mathrm{27}}={a}+{a}+{c}−\frac{{a}}{\mathrm{3}}+\frac{{a}}{\mathrm{9}}−\frac{{c}}{\mathrm{27}} \\ $$$$={a}\left(\mathrm{2}−\frac{\mathrm{2}}{\mathrm{9}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{27}}\right)=\frac{\mathrm{13}}{\mathrm{27}} \\ $$$$\frac{\mathrm{16}}{\mathrm{9}}{a}+\frac{\mathrm{26}}{\mathrm{2}×\mathrm{27}}=\frac{\mathrm{13}}{\mathrm{27}}\Leftrightarrow{a}=\mathrm{0} \\ $$$${a}={b}={e}={d}=\mathrm{0} \\ $$$$\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right) \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right) \\ $$$${let}\:{m}+\mathrm{1}={n}−\mathrm{1} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({m}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right) \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{3}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{9}}{\mathrm{16}} \\ $$
Commented by maxmathsup by imad last updated on 06/Apr/19
sir Smail you answer is correct thanks a lot .
$${sir}\:{Smail}\:{you}\:{answer}\:{is}\:{correct}\:{thanks}\:{a}\:{lot}\:. \\ $$

Pin

Leave a Reply

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