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Question Number 57647 by maxmathsup by imad last updated on 09/Apr/19

find approximate value of ξ(3) by using n−1 ≤n≤n+1 for n integr natural .

$${find}\:\:{approximate}\:{value}\:{of}\:\xi\left(\mathrm{3}\right)\:{by}\:{using}\:\:\:{n}−\mathrm{1}\:\leqslant{n}\leqslant{n}+\mathrm{1}\:\:\:{for}\:{n}\:{integr} \\ $$$${natural}\:. \\ $$

Commented by maxmathsup by imad last updated on 18/Apr/19

let determine S_2 =Σ_(n=2) ^∞ (1/((n−1)n^2 )) +1 ⇒ S_2 =lim_(n→+∞) Σ_(k=1) ^n (1/(k(k+1)^2 )) +1 let decompose G(x) =(1/(x(x+1)^2 )) ⇒G(x) =(a/x) +(b/(x+1)) +(c/((x+1)^2 )) a =lim_(x→+∞) xG(x) =1 , c =lim_(x→−1) (x+1)^2 G(x) = −1 ⇒ G(x) =(1/x) +(b/(x+1)) −(1/((x+1)^2 )) G(1) =(1/4) =1 +(b/2) −(1/4) ⇒1 =4 +2b −1 =3+2b ⇒2b =−2 ⇒b =−1 ⇒ G(x) =(1/x) −(1/(x+1)) −(1/((x+1)^2 )) ⇒Σ_(k=1) ^n G(k) +1 =Σ_(k=1) ^n (1/k) −Σ_(k=1) ^n (1/(k+1)) −Σ_(k=1) ^n (1/((k+1)^2 )) +1 =Σ_(k=1) ^n (1/k) −Σ_(k=2) ^(n+1) (1/k) −Σ_(k=2) ^(n+1) (1/k^2 ) +1 but Σ_(k=1) ^n (1/k) =H_n , Σ_(k=2) ^(n+1) (1/k) =H_(n+1) −1 and Σ_(k=2) ^(n+1) (1/k^2 ) =ξ_(n+1) (2)−1 ⇒ S_2 =H_n −H_(n+1) +1 −ξ_(n+1) (2) +1 →2−(π^2 /6) ⇒ (π^2 /6) −(1/2) ≤ξ(3)≤2−(π^2 /6) ⇒((π^2 −3)/6) ≤ξ(3)≤((12−π^2 )/6)

$${let}\:{determine}\:{S}_{\mathrm{2}} =\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }\:+\mathrm{1}\:\Rightarrow\:{S}_{\mathrm{2}} ={lim}_{{n}\rightarrow+\infty} \sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:+\mathrm{1} \\ $$$${let}\:{decompose}\:{G}\left({x}\right)\:=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{G}\left({x}\right)\:=\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow+\infty} \:{xG}\left({x}\right)\:=\mathrm{1}\:,\:{c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {G}\left({x}\right)\:=\:−\mathrm{1}\:\Rightarrow \\ $$$${G}\left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${G}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\:=\mathrm{1}\:+\frac{{b}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow\mathrm{1}\:=\mathrm{4}\:+\mathrm{2}{b}\:−\mathrm{1}\:=\mathrm{3}+\mathrm{2}{b}\:\Rightarrow\mathrm{2}{b}\:=−\mathrm{2}\:\Rightarrow{b}\:=−\mathrm{1}\:\Rightarrow \\ $$$${G}\left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{G}\left({k}\right)\:+\mathrm{1}\: \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:+\mathrm{1} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:+\mathrm{1}\:\:\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} \:\:,\:\:\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{1}} −\mathrm{1}\:\:\:\:{and}\:\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1}\:\Rightarrow \\ $$$${S}_{\mathrm{2}} ={H}_{{n}} −{H}_{{n}+\mathrm{1}} \:+\mathrm{1}\:−\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)\:+\mathrm{1}\:\:\rightarrow\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\Rightarrow\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\xi\left(\mathrm{3}\right)\leqslant\mathrm{2}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\frac{\pi^{\mathrm{2}} −\mathrm{3}}{\mathrm{6}}\:\leqslant\xi\left(\mathrm{3}\right)\leqslant\frac{\mathrm{12}−\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 18/Apr/19

we have (n−1)n^2 ≤n^3 ≤(n+1)n^2 ⇒(1/((n+1)n^2 )) ≤(1/n^3 ) ≤(1/((n−1)n^2 )) ⇒ Σ_(n=2) ^∞ (1/((n+1)n^2 )) ≤ Σ_(n=2) ^∞ (1/n^3 ) ≤ Σ_(n=2) ^∞ (1/((n−1)n^2 )) ⇒Σ_(n=2) ^∞ (1/((n+1)n^2 )) ≤ξ(3)−1 ≤Σ_(n=2) ^∞ (1/((n−1)n^2 )) ⇒Σ_(n=2) ^∞ (1/((n+1)n^2 )) +1 ≤ξ(3)≤Σ_(n=2) ^∞ (1/((n−1)n^2 )) +1 let determine S_1 =1+Σ_(n=2) ^∞ (1/((n+1)n^2 )) ⇒ S_1 =(1/2) +Σ_(n=1) ^∞ (1/((n+1)n^2 )) S_1 =lim_(n→+∞) S_n with S_n =(1/2) +Σ_(k=1) ^n (1/((k+1)k^2 )) let decompose F(x)=(1/((x+1)x^2 )) F(x) =(a/(x+1)) +(b/x) +(c/x^2 ) a =lim_(x→−1) (x+1)F(x) =1 ,c =lim_(x→0) x^2 F(x) =1 ⇒F(x)=(1/(x+1)) +(b/x) +(1/x^2 ) F(1) =(1/2) =(1/2) +b +1 ⇒b =−1 ⇒F(x) =(1/(x+1)) −(1/x) +(1/x^2 ) ⇒ Σ_(k=1) ^n F(k) =Σ_(k=1) ^n (1/(k+1)) −Σ_(k=1) ^n (1/k) +Σ_(k=1) ^n (1/k^2 ) =Σ_(k=2) ^(n+1) (1/k) −Σ_(k=1) ^n (1/k) +Σ_(k=1) ^n (1/k^2 ) =−1−(1/(n+1)) +Σ_(k=1) ^n (1/k^2 ) →(π^2 /6) −1 ⇒ S_1 =(π^2 /6) −(1/2)

$${we}\:{have}\:\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} \:\leqslant{n}^{\mathrm{3}} \:\leqslant\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} \:\Rightarrow\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} }\:\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\leqslant\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} }\:\leqslant\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\leqslant\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} }\:\leqslant\xi\left(\mathrm{3}\right)−\mathrm{1} \\ $$$$\leqslant\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} }\:+\mathrm{1}\:\leqslant\xi\left(\mathrm{3}\right)\leqslant\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}^{\mathrm{2}} }\:+\mathrm{1} \\ $$$${let}\:{determine}\:{S}_{\mathrm{1}} =\mathrm{1}+\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} }\:\Rightarrow\:{S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} } \\ $$$${S}_{\mathrm{1}} ={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\:{with}\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){k}^{\mathrm{2}} }\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right){x}^{\mathrm{2}} } \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{b}}{{x}}\:+\frac{{c}}{{x}^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\mathrm{1}\:\:\:,{c}\:={lim}_{{x}\rightarrow\mathrm{0}} \:{x}^{\mathrm{2}} \:{F}\left({x}\right)\:=\mathrm{1}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\frac{{b}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+{b}\:+\mathrm{1}\:\Rightarrow{b}\:=−\mathrm{1}\:\Rightarrow{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$$=\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=−\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\rightarrow\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\mathrm{1}\:\Rightarrow \\ $$$${S}_{\mathrm{1}} =\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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