Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 45237 by maxmathsup by imad last updated on 10/Oct/18

calculate Σ_(n=2) ^∞ (1/(n^3 −n))

$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} −{n}} \\ $$

Commented by maxmathsup by imad last updated on 11/Oct/18

let decompose F(x)=(1/(x^3 −x)) =(1/(x(x−1)(x+1))) F(x)=(a/x) +(b/(x−1)) +(c/(x+1)) a =lim_(x→0) xF(x) =−1 b =lim_(x→1) (x−1)F(x)=(1/2) c =lim_(x→−1) (x+1)F(x) =(1/2) ⇒F(x)=−(1/x) +(1/(2(x−1))) +(1/(2(x+1))) let S_n =Σ_(k=2) ^n (1/(k^3 −k)) ⇒S_n =Σ_(k=2) ^n F(k) =−Σ_(k=2) ^n (1/k) +(1/2)Σ_(k=2) ^n (1/(k−1)) +(1/2)Σ_(k=2) ^n (1/(k+1)) but Σ_(k=2) ^n (1/k) =H_n −1 Σ_(k=2) ^n (1/(k−1)) =Σ_(k=1) ^(n−1) (1/k) =H_(n−1) Σ_(k=2) ^n (1/(k+1)) =Σ_(k=3) ^(n+1) (1/k)=H_(n+1) −(3/2) ⇒S_n =−H_(n+1) +1+(1/2) H_(n−1) +(1/2)H_(n+1) −(3/4) ⇒ S_n =(1/2)(ln(n−1)+γ +ln(n+1)+γ +o((1/n)))−ln(n)−γ +o((1/n)) +(1/4) =ln((√(n−1))(√(n+1)))−ln(n)+o((1/n))+(1/4) =ln{((√(n^2 −1))/n)}+o((1/n))+(1/4) →(1/4) (n→+∞) so lim_(n→+∞) S_n =(1/4) .

$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} −{x}}\:=\frac{\mathrm{1}}{{x}\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{{x}}\:+\frac{{b}}{{x}−\mathrm{1}}\:+\frac{{c}}{{x}+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow\mathrm{0}} {xF}\left({x}\right)\:=−\mathrm{1} \\ $$$${b}\:={lim}_{{x}\rightarrow\mathrm{1}} \left({x}−\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${c}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({x}\right)=−\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)} \\ $$$${let}\:{S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{3}} −{k}}\:\Rightarrow{S}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:{F}\left({k}\right)\:=−\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${but}\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}} −\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}−\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{k}}\:={H}_{{n}−\mathrm{1}} \\ $$$$\sum_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{{k}}={H}_{{n}+\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{S}_{{n}} =−{H}_{{n}+\mathrm{1}} \:+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:{H}_{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}+\mathrm{1}} −\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left({n}−\mathrm{1}\right)+\gamma\:+{ln}\left({n}+\mathrm{1}\right)+\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right)−{ln}\left({n}\right)−\gamma\:+{o}\left(\frac{\mathrm{1}}{{n}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$={ln}\left(\sqrt{{n}−\mathrm{1}}\sqrt{{n}+\mathrm{1}}\right)−{ln}\left({n}\right)+{o}\left(\frac{\mathrm{1}}{{n}}\right)+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$={ln}\left\{\frac{\sqrt{{n}^{\mathrm{2}} −\mathrm{1}}}{{n}}\right\}+{o}\left(\frac{\mathrm{1}}{{n}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\:\rightarrow\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\left({n}\rightarrow+\infty\right)\:{so}\:\:{lim}_{{n}\rightarrow+\infty} {S}_{{n}} =\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$

Answered by Smail last updated on 11/Oct/18

(1/(n(n^2 −1)))=(1/(n(n−1)(n+1)))=(a/n)+(b/(n−1))+(c/(n+1)) a=−1 ; b=(1/2) ; c=(1/2) Σ_(n=2) ^∞ (1/(n^3 −n))=Σ_(n=2) ^∞ (−(1/n)+(1/(2(n−1)))+(1/(2(n+1)))) =(1/2)Σ_(n=2) ^∞ (1/(n−1))+(1/2)Σ_(n=2) ^∞ (1/(n+1))−Σ_(n=2) ^∞ (1/n) =(1/2)(1+(1/2)+(1/3)+...)+(1/2)((1/3)+(1/4)+...)−((1/2)+(1/3)+...) =(1/2)(1+(1/2)+(1/3)+(1/3)+(1/4)+(1/4)+...)−((1/2)+(1/3)+...) =(1/2)(1+(1/2))+((1/3)+(1/4)+(1/5)+...)−(1/2)−((1/3)+(1/4)+(1/5)+...) =(1/2)(1+(1/2))−(1/2)=(1/4)

$$\frac{\mathrm{1}}{{n}\left({n}^{\mathrm{2}} −\mathrm{1}\right)}=\frac{\mathrm{1}}{{n}\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}=\frac{{a}}{{n}}+\frac{{b}}{{n}−\mathrm{1}}+\frac{{c}}{{n}+\mathrm{1}} \\ $$$${a}=−\mathrm{1}\:;\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\:;\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} −{n}}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+...\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+...\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+...\right)−\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+...\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by Tawa1 last updated on 11/Oct/18

Sir please what is the sum of nth term of (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ... = ?

$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:+\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:+\:...\:\:=\:\:? \\ $$

Commented by maxmathsup by imad last updated on 11/Oct/18

your answer is correct sir sma3l tbanks..

$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{sma}\mathrm{3}{l}\:\:{tbanks}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com