Menu Close

find-n-1-1-n-2-n-1-3-

Tinku Tara 0 Comments
Pin




Question Number 68592 by Abdo msup. last updated on 14/Sep/19
find Σ_(n=1) ^∞ (1/(n^2 (n+1)^3 ))
$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by Abdo msup. last updated on 16/Sep/19
let S_n =Σ_(k=1) ^n (1/(k^2 (k+1)^3 )) we have S=lim_(n→+∞) S_n 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) +(1/x^2 )−(a/(x+1)) +(d/((x+1)^2 )) +(1/((x+1)^3 )) F(1)=(1/8) =a+1−(a/2) +(d/4) +(1/8) ⇒ (1/2)a +(d/4)=−1 ⇒2a+d =−4 F(−2) =−(1/4) =−(a/2)+(1/4) +a +d−1 =(1/2)a +d−(3/4) ⇒−1=2a +4d−3 ⇒ 2a+4d=2 ⇒a+2d =1 but d=−2a−4 ⇒ a−4a−8 =1 ⇒−3a=9 ⇒a=−3 ⇒d=−2a−4 =6−4=2 F(x)=−(3/x) +(1/x^2 ) +(3/((x+1)))+(2/((x+1)^2 )) +(1/((x+1)^3 )) ⇒ S_n =Σ_(k=1) ^n F(k)=−3Σ_(k=1) ^n (1/k) +Σ_(k=1) ^n (1/k^2 ) +3 Σ_(k=1) ^n (1/(k+1)) +2Σ_(k=1) ^n (1/((k+1)^2 )) +Σ_(k=1) ^n (1/((k+1)^3 )) Σ_(k=1) ^n (1/(k+1)) =Σ_(k=2) ^(n+1) (1/k) =H_(n+1) −1 Σ_(k=1) ^n (1/k^2 ) =ξ_n (2) Σ_(k=1) ^n (1/((k+1)^2 )) =Σ_(k=2) ^(n+1) (1/k^2 ) =ξ_(n+1) (2)−1 Σ_(k=1) ^n (1/((k+1)^3 )) =Σ_(k=2) ^(n+1) (1/k^3 ) =ξ_(n+1) (3)−1 ⇒ S_n =−(3/5)H_n +ξ_n (2)−(3/5)(H_(n+1) −1) −(2/5)(ξ_(n+1) (2)−1) +ξ_(n+1) (3)−1 ⇒ S_n =−3H_n +ξ_n (2) +3(H_(n+1) −1) +2(ξ_(n+1) (2)−1) +ξ_(n+1) (3)−1 =3(H_(n+1) −H_n ) +ξ_n (2)+2ξ_(n+1) (2)+ξ_(n+1) (3)−6 but lim_(n→+∞) (H_(n+1) −H_n )=0 ⇒ lim_(n→+∞) S_n = 3ξ(2)+ξ(3)−6 =3.(π^2 /6) +ξ(3)−6 =(π^2 /2) +ξ(3)−6 with ξ(3)∼1,2
$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{3}} }\:\:{we}\:{have}\:{S}={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{{e}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {F}\left({x}\right)=\mathrm{1} \\ $$$${e}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} {F}\left({x}\right)=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\mathrm{0}={a}\:+{c}\:\Rightarrow{c}=−{a}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{8}}\:={a}+\mathrm{1}−\frac{{a}}{\mathrm{2}}\:+\frac{{d}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{a}\:+\frac{{d}}{\mathrm{4}}=−\mathrm{1}\:\Rightarrow\mathrm{2}{a}+{d}\:=−\mathrm{4} \\ $$$${F}\left(−\mathrm{2}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\:=−\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\:+{a}\:+{d}−\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{a}\:+{d}−\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow−\mathrm{1}=\mathrm{2}{a}\:+\mathrm{4}{d}−\mathrm{3}\:\Rightarrow \\ $$$$\mathrm{2}{a}+\mathrm{4}{d}=\mathrm{2}\:\Rightarrow{a}+\mathrm{2}{d}\:=\mathrm{1}\:\:\:{but}\:{d}=−\mathrm{2}{a}−\mathrm{4}\:\Rightarrow \\ $$$${a}−\mathrm{4}{a}−\mathrm{8}\:=\mathrm{1}\:\Rightarrow−\mathrm{3}{a}=\mathrm{9}\:\Rightarrow{a}=−\mathrm{3} \\ $$$$\Rightarrow{d}=−\mathrm{2}{a}−\mathrm{4}\:=\mathrm{6}−\mathrm{4}=\mathrm{2} \\ $$$${F}\left({x}\right)=−\frac{\mathrm{3}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{\left({x}+\mathrm{1}\right)}+\frac{\mathrm{2}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)=−\mathrm{3}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$+\mathrm{3}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:+\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }\: \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{1}} −\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}} \left(\mathrm{2}\right) \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }\:=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:=\xi_{{n}+\mathrm{1}} \left(\mathrm{3}\right)−\mathrm{1}\:\Rightarrow \\ $$$${S}_{{n}} =−\frac{\mathrm{3}}{\mathrm{5}}{H}_{{n}} \:+\xi_{{n}} \left(\mathrm{2}\right)−\frac{\mathrm{3}}{\mathrm{5}}\left({H}_{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$−\frac{\mathrm{2}}{\mathrm{5}}\left(\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{3}\right)−\mathrm{1}\:\Rightarrow \\ $$$${S}_{{n}} =−\mathrm{3}{H}_{{n}} \:+\xi_{{n}} \left(\mathrm{2}\right)\:+\mathrm{3}\left({H}_{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$+\mathrm{2}\left(\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{3}\right)−\mathrm{1} \\ $$$$=\mathrm{3}\left({H}_{{n}+\mathrm{1}} −{H}_{{n}} \right)\:+\xi_{{n}} \left(\mathrm{2}\right)+\mathrm{2}\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)+\xi_{{n}+\mathrm{1}} \left(\mathrm{3}\right)−\mathrm{6} \\ $$$${but}\:{lim}_{{n}\rightarrow+\infty} \:\left({H}_{{n}+\mathrm{1}} −{H}_{{n}} \right)=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\:\mathrm{3}\xi\left(\mathrm{2}\right)+\xi\left(\mathrm{3}\right)−\mathrm{6} \\ $$$$=\mathrm{3}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\xi\left(\mathrm{3}\right)−\mathrm{6}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:+\xi\left(\mathrm{3}\right)−\mathrm{6}\:\:{with} \\ $$$$\xi\left(\mathrm{3}\right)\sim\mathrm{1},\mathrm{2} \\ $$
Commented by Abdo msup. last updated on 16/Sep/19
ξ_n (x)=Σ_(k=1) ^n (1/k^x )
$$\xi_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{{x}} } \\ $$

Pin

Leave a Reply

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