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Question Number 28814 by abdo imad last updated on 30/Jan/18
find Σ_(n=1) ^∞ (1/(1^3 +2^3 +...+n^3 )) .
$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} \:+\mathrm{2}^{\mathrm{3}} +…+{n}^{\mathrm{3}} }\:. \\ $$
Commented by abdo imad last updated on 01/Feb/18
let put S_n = Σ_(k=1) ^n (1/(1^3 +2^3 +....+k^3 )) and S=Σ_(n=1) ^∞ (1/(1^3 +2^3 +...+n^3 )) =lim_(n→+∞) S_n we know that 1^(3 ) +2^3 +....k^3 = (((k(k+1))/2))^2 = ((k^2 (k+1)^2 )/4) so S_n = 4 Σ_(k=1) ^n (1/(k^2 (k+1)^2 )) let decompose F(x)= (1/(x^2 (x+1)^2 )) F(x)= (a/x) +(b/x^2 ) +(c/(x+1)) +(d/((x+1)^2 )) b=lim_(x→0) x^2 f(x)= 1 , d=lim_(x→−1) (x+1)^2 F(x)= 1 so F(x)= (a/x) +(1/x^2 ) +(c/(x+1)) + (1/((x+1)^2 )) lim _(x→+∞) xF(x)= 0= a+c ⇒c=−a so F(x)= (a/x) −(a/(x+1)) + (1/x^2 ) + (1/((x+1)^2 )) F(1)= (1/4)=a −(a/2) + +1 +(1/4)⇒(a/2) +1=0 ⇒a=−2 so F(x)= ((−2)/x) +(2/(x+1)) + (1/x^2 ) + (1/((x+1)^2 )) Σ_(k=1) ^n F(k)= −2Σ_(k=1) ^n (1/k) + 2 Σ_(k=1) ^n (1/(k+1)) + Σ_(k=1) ^n (1/k^2 ) +Σ_(k=1) ^n (1/((k+1)^2 )) but Σ_(k=1) ^n (1/k)=H_n 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) with ξ_n (x)=Σ_(k=1) ^n (1/k^x ) . 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 F(k)= −2H_n +2H_(n+1) −2 +ξ_n (2) +ξ_(n+1) (2)−1 =2(H_(n+1) −H_n ) +ξ_n (2) +ξ_(n+1) (2)−1 →0 +(π^2 /6) +(π^2 /6)−1= (π^2 /3) −1 and lim_(n→+∞) S_n =4( (π^2 /3)−1)=S .
$${let}\:{put}\:\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} \:+\mathrm{2}^{\mathrm{3}} +….+{k}^{\mathrm{3}} }\:\:\:{and}\:{S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} \:+\mathrm{2}^{\mathrm{3}} +…+{n}^{\mathrm{3}} }\:={lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\:{we}\:{know}\:{that} \\ $$$$\mathrm{1}^{\mathrm{3}\:} \:+\mathrm{2}^{\mathrm{3}} \:+….{k}^{\mathrm{3}} =\:\left(\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}\right)^{\mathrm{2}} =\:\frac{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\:{so} \\ $$$${S}_{{n}} =\:\mathrm{4}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:{let}\:{decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\frac{{d}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${b}={lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} {f}\left({x}\right)=\:\mathrm{1}\:\:,\:{d}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} {F}\left({x}\right)=\:\mathrm{1}\:{so} \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}\:_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\:\mathrm{0}=\:{a}+{c}\:\Rightarrow{c}=−{a}\:{so} \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:−\frac{{a}}{{x}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{4}}={a}\:−\frac{{a}}{\mathrm{2}}\:+\:+\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\frac{{a}}{\mathrm{2}}\:+\mathrm{1}=\mathrm{0}\:\Rightarrow{a}=−\mathrm{2}\:{so} \\ $$$${F}\left({x}\right)=\:\frac{−\mathrm{2}}{{x}}\:+\frac{\mathrm{2}}{{x}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)=\:−\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}}\:+\:\mathrm{2}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:+\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\: \\ $$$$+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{but} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}=\:\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\:\frac{\mathrm{1}}{{k}}=\:{H}_{{n}+\mathrm{1}} −\mathrm{1}\:\: \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\:\xi_{{n}} \left(\mathrm{2}\right)\:\:\:\:{with}\:\xi_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{{x}} }\:. \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\:\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)=\:−\mathrm{2}{H}_{{n}} \:+\mathrm{2}{H}_{{n}+\mathrm{1}} −\mathrm{2}\:+\xi_{{n}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1} \\ $$$$=\mathrm{2}\left({H}_{{n}+\mathrm{1}} −{H}_{{n}} \right)\:+\xi_{{n}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\rightarrow\mathrm{0}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{1}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{1} \\ $$$${and}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\mathrm{4}\left(\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}−\mathrm{1}\right)={S}\:. \\ $$
Commented by abdo imad last updated on 01/Feb/18
Σ_(k=1) ^n F(k)= 2(H_(n+1) −H_n ) +ξ_n (2) +ξ_(n+1) (2)−3 →0 +((2π^2 )/6) −3= (π^2 /3)−3 and lim_(n→+∞) S_n =4( (π^2 /3) −3) = ((4π^2 )/3) −12.
$$\sum_{{k}=\mathrm{1}} ^{{n}} {F}\left({k}\right)=\:\mathrm{2}\left({H}_{{n}+\mathrm{1}} −{H}_{{n}} \right)\:+\xi_{{n}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{3} \\ $$$$\rightarrow\mathrm{0}\:+\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{6}}\:−\mathrm{3}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}−\mathrm{3}\:\:\:\:{and} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\mathrm{4}\left(\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{3}\right)\:=\:\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{12}. \\ $$

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