Menu Close

find-the-value-of-n-1-1-n-1-n-2-n-3-in-terms-of-3-we-give-x-n-1-1-n-x-and-x-gt-1-zeta-function-of-Rieman-

Tinku Tara 0 Comments
Pin




Question Number 26222 by abdo imad last updated on 22/Dec/17
find the value of Σ_(n=1) ^∝ (1/((n+1)(n+2)n^3 )) in terms of ξ(3) we give ξ(x)= Σ_(n=1) ^∝ (1/n^x ) and x>1 (zeta function of Rieman)
$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right){n}^{\mathrm{3}} }\:{in}\:{terms}\:{of}\:\xi\left(\mathrm{3}\right) \\ $$$${we}\:{give}\:\xi\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{and}\:{x}>\mathrm{1} \\ $$$$\left({zeta}\:{function}\:{of}\:{Rieman}\right) \\ $$
Commented by abdo imad last updated on 24/Dec/17
we decompose the fraction F(x)= (1/((x+1)(x+2) x^3 )) in simple elements F(x) = (a/(x+1)) + (b/(x+2)) + (c/x) + (d/x^2 ) + (e/x^3 ) a= lim_(x−>−1) (x+1)F(x)= −1 b= lim_(x−>−2 ) (x+2)F(x) = (1/8) e=lim_(x−>0) x^3 F(x)= (1/2) ⇒ F(x)= −(1/(x+1))+ (1/(8(x+2)))+ (c/x) +(d/x^2 ) + (1/(2x^3 )) lim_(x−>∝) xF(x)=0=−1+ (1/8) +c ⇒c= (7/8) F(x)= −(1/(x+1)) + (1/(8(x+2))) + (7/(8x)) + (d/x^2 ) + (1/(2x^3 )) F(1)= (1/6) = −(1/2) + (1/(24)) + (7/8) +d + (1/2) = ((11)/(12)) +d ⇒d= (1/6) −((11)/(12))=−(3/4) F(x)= −(1/(x+1)) + (1/(8(x+2))) + (7/(8x)) − (3/(4x^2 )) + (1/(2x^3 )) let put S_n = Σ_(k=1) ^(k=n) (1/((n+1)(n+2) n^3 )) S_n = −Σ_(k=1) ^(k=n) (1/(k+1))+ (1/8) Σ_(k=1) ^(k=n) (1/(k+2)) +(7/8)Σ_(k=1) ^(k=n) (1/k) −(3/4) Σ_(k=1) ^(k=n) (1/k^2 ) +(1/2) Σ_(k=1) ^(k=n) (1/n^3 ) but Σ_(k=1) ^(k=n) (1/(k+1))= Σ_(k=2) ^(k=n+1) (1/k) = H_(n+1) −1 Σ_(k=1) ^(k=n) (1/k) =H_n Σ_(k=1) ^(k=n) (1/(k+2)) = Σ_(k=3) ^(n+2) (1/k) = H_(n+2) −(3/2) S_n = −H_(n+1) +1 +(1/8) H_(n+2) −(3/(16)) + (7/8) H_n −(3/4) ξ_n (2) + (1/2)ξ_n (3) = ((13)/(16 )) + (1/8) H_(n+2) +(7/8) H_n − H_(n+1) −(3/4) ξ_n (2) + (1/2)ξ_n (3) but (1/8) H_(n+2) +(7/8) H_n −H_(n+1) =(1/8)( ln(n+2) +γ +o((1/n)))+(7/8)(ln(n)+γ+o((1/n))−( ln(n+1)+γ+o((1/n))) =ln( (n+2)^(1/8) .n^(7/8) )−ln(n+1) +o((1/(n)))) = ln( (((n+2)^(1/8) .n^(7/8) )/(n+1)) )−−−>0_(n−>∝) lim_(n−>∝ ) ξ_n (2)=ξ(2)and lim_(n−>∝) ξ_n (3)=ξ(3)⇒ lim_(n−>∝ ) S_n = ((13)/(16)) +(1/2) ξ(3) −(3/4) ξ(2)
$${we}\:{decompose}\:{the}\:{fraction}\:{F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\:{x}^{\mathrm{3}} }\:{in}\:{simple} \\ $$$${elements}\:\:{F}\left({x}\right)\:\:=\:\frac{{a}}{{x}+\mathrm{1}}\:+\:\frac{{b}}{{x}+\mathrm{2}}\:+\:\frac{{c}}{{x}}\:+\:\frac{{d}}{{x}^{\mathrm{2}} }\:+\:\frac{{e}}{{x}^{\mathrm{3}} } \\ $$$${a}=\:{lim}_{{x}−>−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\:−\mathrm{1} \\ $$$${b}=\:{lim}_{{x}−>−\mathrm{2}\:} \left({x}+\mathrm{2}\right){F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${e}={lim}_{{x}−>\mathrm{0}} {x}^{\mathrm{3}} {F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\Rightarrow\:{F}\left({x}\right)=\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}+\:\:\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{2}\right)}+\:\frac{{c}}{{x}}\:+\frac{{d}}{{x}^{\mathrm{2}} }\:\:+\:\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$${lim}_{{x}−>\propto} {xF}\left({x}\right)=\mathrm{0}=−\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{8}}\:+{c}\:\:\Rightarrow{c}=\:\frac{\mathrm{7}}{\mathrm{8}} \\ $$$${F}\left({x}\right)=\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{2}\right)}\:+\:\frac{\mathrm{7}}{\mathrm{8}{x}}\:+\:\frac{{d}}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$${F}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{6}}\:\:=\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{24}}\:\:+\:\frac{\mathrm{7}}{\mathrm{8}}\:\:+{d}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:=\:\frac{\mathrm{11}}{\mathrm{12}}\:\:+{d}\:\:\Rightarrow{d}=\:\frac{\mathrm{1}}{\mathrm{6}}\:−\frac{\mathrm{11}}{\mathrm{12}}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${F}\left({x}\right)=\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{8}\left({x}+\mathrm{2}\right)}\:+\:\frac{\mathrm{7}}{\mathrm{8}{x}}\:\:−\:\frac{\mathrm{3}}{\mathrm{4}{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\:{n}^{\mathrm{3}} } \\ $$$${S}_{{n}} \:\:=\:−\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}+\:\frac{\mathrm{1}}{\mathrm{8}}\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:+\frac{\mathrm{7}}{\mathrm{8}}\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}}\:\:−\frac{\mathrm{3}}{\mathrm{4}}\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$${but}\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}=\:\sum_{{k}=\mathrm{2}} ^{{k}={n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}}\:\:=\:{H}_{{n}+\mathrm{1}} \:\:−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}}\:\:={H}_{{n}} \\ $$$$\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:=\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \frac{\mathrm{1}}{{k}}\:\:=\:{H}_{{n}+\mathrm{2}} \:\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${S}_{{n}} =\:−{H}_{{n}+\mathrm{1}} \:+\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{8}}\:{H}_{{n}+\mathrm{2}} \:\:−\frac{\mathrm{3}}{\mathrm{16}}\:\:+\:\frac{\mathrm{7}}{\mathrm{8}}\:\:{H}_{{n}} \:−\frac{\mathrm{3}}{\mathrm{4}}\:\xi_{{n}} \left(\mathrm{2}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\xi_{{n}} \left(\mathrm{3}\right) \\ $$$$=\:\frac{\mathrm{13}}{\mathrm{16}\:}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:\:{H}_{{n}+\mathrm{2}} \:\:+\frac{\mathrm{7}}{\mathrm{8}}\:\:{H}_{{n}} \:−\:{H}_{{n}+\mathrm{1}} \:\:−\frac{\mathrm{3}}{\mathrm{4}}\:\xi_{{n}} \left(\mathrm{2}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\xi_{{n}} \left(\mathrm{3}\right) \\ $$$${but}\:\frac{\mathrm{1}}{\mathrm{8}}\:{H}_{{n}+\mathrm{2}} \:\:+\frac{\mathrm{7}}{\mathrm{8}}\:{H}_{{n}} \:\:−{H}_{{n}+\mathrm{1}} \:\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\:\:{ln}\left({n}+\mathrm{2}\right)\:+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)+\frac{\mathrm{7}}{\mathrm{8}}\left({ln}\left({n}\right)+\gamma+{o}\left(\frac{\mathrm{1}}{{n}}\right)−\left(\:{ln}\left({n}+\mathrm{1}\right)+\gamma+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)\right. \\ $$$$={ln}\left(\:\:\left({n}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} \:.{n}^{\frac{\mathrm{7}}{\mathrm{8}}} \:\:\right)−{ln}\left({n}+\mathrm{1}\right)\:+{o}\left(\frac{\mathrm{1}}{\left.{n}\right)}\right) \\ $$$$=\:{ln}\left(\:\:\frac{\left({n}+\mathrm{2}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} \:.{n}^{\frac{\mathrm{7}}{\mathrm{8}}} }{{n}+\mathrm{1}}\:\:\right)−−−>\mathrm{0}_{{n}−>\propto} \\ $$$${lim}_{{n}−>\propto\:} \xi_{{n}} \left(\mathrm{2}\right)=\xi\left(\mathrm{2}\right){and}\:{lim}_{{n}−>\propto} \xi_{{n}} \:\left(\mathrm{3}\right)=\xi\left(\mathrm{3}\right)\Rightarrow \\ $$$${lim}_{{n}−>\propto\:} {S}_{{n}} \:\:=\:\frac{\mathrm{13}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\xi\left(\mathrm{3}\right)\:−\frac{\mathrm{3}}{\mathrm{4}}\:\xi\left(\mathrm{2}\right) \\ $$
Commented by abdo imad last updated on 25/Dec/17
ξ_n (x)=Σ_(k=1) ^(k=n) (1/k^x ) with x>1
$$\xi_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}^{{x}} }\:\:{with}\:\:{x}>\mathrm{1} \\ $$

Pin

Leave a Reply

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