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Question Number 46617 by maxmathsup by imad last updated on 29/Oct/18
calculate Σ_(n=1) ^∞ (1/(n(n+1)(n+2)(n+3)(n+4)(n+5)))
$${calculate}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\left({n}+\mathrm{4}\right)\left({n}+\mathrm{5}\right)} \\ $$
Commented by maxmathsup by imad last updated on 29/Oct/18
let decompose F(x)=(1/(x(x+1)(x+2)(x+3)(x+4)(x+5))) F(x)=(a/(x )) +(b/(x+1)) +(c/(x+2)) +(d/(x+3)) +(e/(x+4)) +(f/(x+5)) a =lim_(x→0) xF(x)=(1/(1.2.3.4.5)) =(1/(120)) b=lim_(x→−1) (x+1)F(x)=(1/((−1)1.2.3.4)) =((−1)/(24)) c=lim_(x→−2) (x+2)F(x)= (1/((−2).(−1)1.2.3)) =(1/(12)) d =lim_(x→−3) (x+3)F(x)=(1/((−3)(−2)(−1)1.2)) =((−1)/(12)) e =lim_(x→−4) (x+4)F(x)=(1/((−4)(−3)(−2)(−1).1)) =(1/(24)) f =lim_(x→−5) (x+5)F(x) =(1/((−5)(−4)(−3)(−2)(−1))) =((−1)/(120)) ⇒ F(x)=(1/(120x)) −(1/(24(x+1))) +(1/(12(x+2))) −(1/(12(x+3))) +(1/(24(x+4))) −(1/(120(x+5))) ⇒ Σ_(k=1) ^n F(k) =(1/(120)) Σ_(k=1) ^n (1/k) −(1/(24)) Σ_(k=1) ^n (1/(k+1)) +(1/(12))Σ_(k=1) ^n (1/(k+2)) −(1/(12))Σ_(k=1) ^n (1/(k+3)) +(1/(24)) Σ_(k=1) ^n (1/(k+4)) −(1/(120))Σ_(k=1) ^n (1/(k+5)) =(1/(120)) H_n −(1/(24)){ H_(n+1) −1} +(1/(12)){H_(n+2) −(3/2)} −(1/(12)){H_(n+3) −(3/2)−(1/3)} +(1/(24)){H_(n+4) −(3/2)−(1/3)−(1/4)}−(1/(120)){H_(n+5) −(3/2)−(1/3)−(1/4) −(1/5)} =(1/(120)){H_n −H_(n+5) } +(1/(24)){H_(n+4) −H_(n+1) } +(1/(12)){ H_(n+2) −H_(n+3) } +(1/(24)) −(1/8) +(1/(12))((3/2)+(1/3))−(1/(24))((3/2)+(1/3)+(1/4))+(1/(120)){(3/2) +(1/3) +(1/4) +(1/5)} but H_n −H_(n+5) →0 ,H_(n+4) −H_(n+1) →0 ^� H_(n+2) −H_(n+3) →0 (n→+∞) after reducton of calculus we get lim_(n→+∞) Σ_(k=1) ^n F(k) =(1/(600)) =S .
$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{4}\right)\left({x}+\mathrm{5}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}\:}\:+\frac{{b}}{{x}+\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{2}}\:+\frac{{d}}{{x}+\mathrm{3}}\:+\frac{{e}}{{x}+\mathrm{4}}\:+\frac{{f}}{{x}+\mathrm{5}} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} {xF}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}}\:=\frac{\mathrm{1}}{\mathrm{120}} \\ $$$${b}={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\left(−\mathrm{1}\right)\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}}\:=\frac{−\mathrm{1}}{\mathrm{24}} \\ $$$${c}={lim}_{{x}\rightarrow−\mathrm{2}} \left({x}+\mathrm{2}\right){F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(−\mathrm{2}\right).\left(−\mathrm{1}\right)\mathrm{1}.\mathrm{2}.\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${d}\:={lim}_{{x}\rightarrow−\mathrm{3}} \left({x}+\mathrm{3}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\left(−\mathrm{3}\right)\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)\mathrm{1}.\mathrm{2}}\:=\frac{−\mathrm{1}}{\mathrm{12}} \\ $$$${e}\:={lim}_{{x}\rightarrow−\mathrm{4}} \left({x}+\mathrm{4}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\left(−\mathrm{4}\right)\left(−\mathrm{3}\right)\left(−\mathrm{2}\right)\left(−\mathrm{1}\right).\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${f}\:={lim}_{{x}\rightarrow−\mathrm{5}} \left({x}+\mathrm{5}\right){F}\left({x}\right)\:=\frac{\mathrm{1}}{\left(−\mathrm{5}\right)\left(−\mathrm{4}\right)\left(−\mathrm{3}\right)\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)}\:=\frac{−\mathrm{1}}{\mathrm{120}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{120}{x}}\:−\frac{\mathrm{1}}{\mathrm{24}\left({x}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{12}\left({x}+\mathrm{2}\right)}\:−\frac{\mathrm{1}}{\mathrm{12}\left({x}+\mathrm{3}\right)}\:+\frac{\mathrm{1}}{\mathrm{24}\left({x}+\mathrm{4}\right)}\:−\frac{\mathrm{1}}{\mathrm{120}\left({x}+\mathrm{5}\right)}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} {F}\left({k}\right)\:=\frac{\mathrm{1}}{\mathrm{120}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}}\:−\frac{\mathrm{1}}{\mathrm{24}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{12}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{12}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{3}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{24}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{120}}\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}+\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{120}}\:{H}_{{n}} −\frac{\mathrm{1}}{\mathrm{24}}\left\{\:{H}_{{n}+\mathrm{1}} −\mathrm{1}\right\}\:+\frac{\mathrm{1}}{\mathrm{12}}\left\{{H}_{{n}+\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\right\}\:−\frac{\mathrm{1}}{\mathrm{12}}\left\{{H}_{{n}+\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{24}}\left\{{H}_{{n}+\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right\}−\frac{\mathrm{1}}{\mathrm{120}}\left\{{H}_{{n}+\mathrm{5}} −\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{5}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{120}}\left\{{H}_{{n}} −{H}_{{n}+\mathrm{5}} \right\}\:+\frac{\mathrm{1}}{\mathrm{24}}\left\{{H}_{{n}+\mathrm{4}} −{H}_{{n}+\mathrm{1}} \right\}\:+\frac{\mathrm{1}}{\mathrm{12}}\left\{\:{H}_{{n}+\mathrm{2}} −{H}_{{n}+\mathrm{3}} \right\} \\ $$$$+\frac{\mathrm{1}}{\mathrm{24}}\:−\frac{\mathrm{1}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{12}}\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{24}}\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{120}}\left\{\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{5}}\right\}\:{but} \\ $$$${H}_{{n}} −{H}_{{n}+\mathrm{5}} \rightarrow\mathrm{0}\:,{H}_{{n}+\mathrm{4}} −{H}_{{n}+\mathrm{1}} \rightarrow\mathrm{0}\bar {\:}{H}_{{n}+\mathrm{2}} −{H}_{{n}+\mathrm{3}} \rightarrow\mathrm{0}\:\left({n}\rightarrow+\infty\right) \\ $$$${after}\:{reducton}\:{of}\:{calculus}\:{we}\:{get}\: \\ $$$${lim}_{{n}\rightarrow+\infty} \sum_{{k}=\mathrm{1}} ^{{n}} \:{F}\left({k}\right)\:=\frac{\mathrm{1}}{\mathrm{600}}\:\:={S}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18
first calculate S_n then the value when n→∞ S_n =C−(1/((n+1)(n+2)(n+3)(n+4)(n+5)×1×5)) S_1 =T_1 =(1/(1×2×3×4×5×6))=(1/(720)) so S_n =C−(1/((n+1)(n+2)(n+3)(n+4)(n+5)×1×5)) (1/(720))=C−(1/(2×3×4×5×6×1×5)) C=(1/(720))+(1/(720×5))=(1/(720))(1+(1/5))=(1/(720))×(6/5) C=(1/(600)) s0 S_n =(1/(600))−(1/((n+1)(n+2)(n+3)(n+4)(n+5)×1×5)) when n→∞ S_n →(1/(600)) so reauired answer is (1/(600))
$${first}\:{calculate}\:{S}_{{n}} \:\:\:{then}\:{the}\:{value}\:{when}\:{n}\rightarrow\infty \\ $$$${S}_{{n}} ={C}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\left({n}+\mathrm{4}\right)\left({n}+\mathrm{5}\right)×\mathrm{1}×\mathrm{5}} \\ $$$${S}_{\mathrm{1}} ={T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{720}} \\ $$$${so}\:{S}_{{n}} ={C}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\left({n}+\mathrm{4}\right)\left({n}+\mathrm{5}\right)×\mathrm{1}×\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{\mathrm{720}}={C}−\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}×\mathrm{6}×\mathrm{1}×\mathrm{5}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{720}}+\frac{\mathrm{1}}{\mathrm{720}×\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{720}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}\right)=\frac{\mathrm{1}}{\mathrm{720}}×\frac{\mathrm{6}}{\mathrm{5}} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{600}} \\ $$$${s}\mathrm{0}\:\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{600}}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\left({n}+\mathrm{4}\right)\left({n}+\mathrm{5}\right)×\mathrm{1}×\mathrm{5}} \\ $$$${when}\:{n}\rightarrow\infty\:\:{S}_{{n}} \rightarrow\frac{\mathrm{1}}{\mathrm{600}} \\ $$$${so}\:{reauired}\:{answer}\:{is}\:\frac{\mathrm{1}}{\mathrm{600}}\:\: \\ $$$$ \\ $$

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