Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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.

$$\sum_{{k}=\mathrm{1}} ^{{n}} {F}\left({k}\right)=\:\mathrm{2}\left({H}_{{n}+\mathrm{1}} −{H}_{{n}} \right)\:+\xi_{{n}} \left(\mathrm{2}\right)\:+\xi_{{n}+\mathrm{1}} \left(\mathrm{2}\right)−\mathrm{3} \\ $$$$\rightarrow\mathrm{0}\:+\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{6}}\:−\mathrm{3}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}−\mathrm{3}\:\:\:\:{and} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\mathrm{4}\left(\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{3}\right)\:=\:\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{3}}\:−\mathrm{12}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com