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Question Number 26877 by abdo imad last updated on 30/Dec/17

find the value of Σ_(n=1) ^∝ (1/(n(n+1)(n+2)(n+3))) .

$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)}\:\:. \\ $$

Commented by abdo imad last updated on 31/Dec/17

let decompose F(x)= (1/(x(x+1)(x+2)(x+3))) F(x)= (a/x) +(b/(x+1)) +(c/(x+2)) +(d/(x+3)) a=lim_(x−>0) xF(x)=(1/6),b= lim_(x−>−1) (x+1)F(x)=−(1/2) c= lim_(x−>−2) (x+2)F(x)= (1/2) , d= lim_(x−>−3) (x+3)F(x)= −(1/6) ⇒ F(x) = (1/(6x)) − (1/(2(x+1))) + (1/(2(x+2))) −(1/(6(x+3))) let put S_n = Σ_(k=1) ^n (1/(k(k+1)(k+2)(k+3))) S_n = (1/6) Σ_(k=1) ^n (1/k)−(1/2) Σ_(k=1) ^n (1/(k+1)) +(1/2) Σ_(k=1) ^(k=n) (1/(k+2)) −(1/6) Σ_(k=1) ^n (1/(k+3)) but Σ_(k=1) ^n (1/k)=H_n Σ_(k=1) ^(k=n) (1/(k+1))= Σ_(k=2) ^(n+1) (1/k) =H_(n+1) −1 Σ_(k=1) ^(k=n) (1/(k+2))= Σ_(k=3) ^(n+2) (1/k)= H_(n+2) −(3/2) Σ_(k=1) ^(k=n) (1/(k+3))= Σ_(k=4) ^(n+3) (1/k) = H_(n+3) −((11)/6) S_n =(1/6) H_n −(1/2)( H_(n+1) −1) +(1/2)( H_(n+2) −(3/2))−(1/6)(H_(n+3) −((11)/6)) =(1/6)(H_n −H_(n+3) )+(1/2)( H_(n+2) −H_(n+1) ) +(1/2)−(3/4)+((11)/(36)) but lim_(n−>∝) (H_n −H_(n+3) )=0 and lim_(n−>∝) ( H_(n+2) −H_(n+1) )=0 ⇒lim_(n−>∝) S_n = −(1/4) +((11)/(36))=((−9+11)/(36))= (2/(36)) =(1/(18)) .

$${let}\:{decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)} \\ $$$${F}\left({x}\right)=\:\frac{{a}}{{x}}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{2}}\:+\frac{{d}}{{x}+\mathrm{3}} \\ $$$${a}={lim}_{{x}−>\mathrm{0}} {xF}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{6}},{b}=\:{lim}_{{x}−>−\mathrm{1}} \:\left({x}+\mathrm{1}\right){F}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}=\:{lim}_{{x}−>−\mathrm{2}} \:\:\left({x}+\mathrm{2}\right){F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:,\:\:{d}=\:{lim}_{{x}−>−\mathrm{3}} \:\left({x}+\mathrm{3}\right){F}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\:{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{6}{x}}\:−\:\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\mathrm{6}\left({x}+\mathrm{3}\right)} \\ $$$${let}\:{put}\:\:{S}_{{n}} \:\:=\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)} \\ $$$$\:{S}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{6}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{6}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}+\mathrm{3}} \\ $$$$ \\ $$$${but}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}={H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \frac{\mathrm{1}}{{k}+\mathrm{1}}=\:\sum_{{k}=\mathrm{2}} ^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{k}}\:={H}_{{n}+\mathrm{1}} \:\:−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \frac{\mathrm{1}}{{k}+\mathrm{2}}=\:\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{{k}}=\:{H}_{{n}+\mathrm{2}} \:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \frac{\mathrm{1}}{{k}+\mathrm{3}}=\:\sum_{{k}=\mathrm{4}} ^{{n}+\mathrm{3}} \:\frac{\mathrm{1}}{{k}}\:=\:{H}_{{n}+\mathrm{3}} \:−\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${S}_{{n}} \:=\frac{\mathrm{1}}{\mathrm{6}}\:{H}_{{n}} \:\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\:{H}_{{n}+\mathrm{1}} \:−\mathrm{1}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\:{H}_{{n}+\mathrm{2}} \:−\frac{\mathrm{3}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{6}}\left({H}_{{n}+\mathrm{3}} −\frac{\mathrm{11}}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left({H}_{{n}} \:−{H}_{{n}+\mathrm{3}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\:{H}_{{n}+\mathrm{2}} −{H}_{{n}+\mathrm{1}} \right)\:+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{11}}{\mathrm{36}} \\ $$$${but}\:{lim}_{{n}−>\propto} \left({H}_{{n}} −{H}_{{n}+\mathrm{3}} \right)=\mathrm{0}\:{and}\:{lim}_{{n}−>\propto} \left(\:{H}_{{n}+\mathrm{2}} \:−{H}_{{n}+\mathrm{1}} \right)=\mathrm{0} \\ $$$$\Rightarrow{lim}_{{n}−>\propto} {S}_{{n}} =\:−\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{11}}{\mathrm{36}}=\frac{−\mathrm{9}+\mathrm{11}}{\mathrm{36}}=\:\frac{\mathrm{2}}{\mathrm{36}}\:=\frac{\mathrm{1}}{\mathrm{18}}\:. \\ $$$$ \\ $$

Commented by abdo imad last updated on 31/Dec/17

this is the method of H_n

$${this}\:{is}\:{the}\:{method}\:{of}\:{H}_{{n}} \\ $$

Answered by prakash jain last updated on 31/Dec/17

(1/(n(n+1)(n+2)(n+3))) =(A/n)+(B/(n+1))+(C/(n+2))+(D/(n+3)) 1=A(n+1)(n+2)(n+3)+Bn(n+2)(n+3) +Cn(n+1)(n+3)+Dn(n+1)(n+2) n=0⇒A=(1/6) n=−1⇒B=−(1/2) n=−2⇒C=(1/2) n=−3⇒D=−(1/6) =(1/(6n))−(1/(2(n+1)))+(1/(2(n+2)))−(1/(6(n+3))) (1/(6.1))−(1/(2∙2))+(1/(2∙3))−(1/(6∙4)) (1/(6.2))−(1/(2∙3))+(1/(2∙4))−(1/(6∙5)) (1/(6.3))−(1/(2∙4))+(1/(2∙5))−(1/(6∙6)) (1/(6.4))−(1/(2∙5))+(1/(2∙6))−(1/(6∙7)) (1/(6(m−2)))−(2/(2(m−1)))+(1/(2(m)))−(1/(6(m+1))) (1/(6(m−1)))−(2/(2(m)))+(1/(2(m+1)))−(1/(6(m+2))) (1/(6(m)))−(2/(2(m+1)))+(1/(2(m+2)))−(1/(6(m+3))) sum to m term (1/(6.1))+(1/(6.2))+(1/(6.3))−(1/(2.2)) + (1/(2(m+2)))−(1/(6(m+1)))−(1/(6(m+2)))−(1/(6(m+3))) =(1/(18))+((3(m+1)(m+3)−(m+2)(m+3)−(m+1)(m+3)−(m+1)(m+2))/(6(m+1)(m+2)(m+3))) =(1/(18))+((2(m+1)(m+3)−(m+2)(m+3)−(m+1)(m+2))/(6(m+1)(m+2)(m+3))) =(1/(18))−(1/(3(m+1)(m+2)(m+3))) sum to infinity= (1/(6∙1))+(1/(6.2))+(1/(6.3))−(1/(2.2))=((6+3+2−9)/(6∙2∙3))=(1/(18))

$$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)} \\ $$$$=\frac{{A}}{{n}}+\frac{{B}}{{n}+\mathrm{1}}+\frac{{C}}{{n}+\mathrm{2}}+\frac{{D}}{{n}+\mathrm{3}} \\ $$$$\mathrm{1}={A}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)+{Bn}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:+{Cn}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{3}\right)+{Dn}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$${n}=\mathrm{0}\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${n}=−\mathrm{1}\Rightarrow{B}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${n}=−\mathrm{2}\Rightarrow{C}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${n}=−\mathrm{3}\Rightarrow{D}=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}{n}}−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({n}+\mathrm{3}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}.\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}.\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}.\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}.\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{7}} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{6}\left({m}−\mathrm{2}\right)}−\frac{\mathrm{2}}{\mathrm{2}\left({m}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({m}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({m}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}\left({m}−\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{2}\left({m}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({m}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({m}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}\left({m}\right)}−\frac{\mathrm{2}}{\mathrm{2}\left({m}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({m}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({m}+\mathrm{3}\right)} \\ $$$$\mathrm{sum}\:\mathrm{to}\:{m}\:\mathrm{term} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}.\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}.\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}.\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{2}} \\ $$$$+\:\frac{\mathrm{1}}{\mathrm{2}\left({m}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({m}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({m}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({m}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}+\frac{\mathrm{3}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{3}\right)−\left({m}+\mathrm{2}\right)\left({m}+\mathrm{3}\right)−\left({m}+\mathrm{1}\right)\left({m}+\mathrm{3}\right)−\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)}{\mathrm{6}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)\left({m}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}+\frac{\mathrm{2}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{3}\right)−\left({m}+\mathrm{2}\right)\left({m}+\mathrm{3}\right)−\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)}{\mathrm{6}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)\left({m}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}−\frac{\mathrm{1}}{\mathrm{3}\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)\left({m}+\mathrm{3}\right)} \\ $$$$ \\ $$$$\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity}= \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}.\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}.\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}.\mathrm{2}}=\frac{\mathrm{6}+\mathrm{3}+\mathrm{2}−\mathrm{9}}{\mathrm{6}\centerdot\mathrm{2}\centerdot\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{18}} \\ $$

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