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Question Number 90787 by Maclaurin Stickker last updated on 26/Apr/20

Find the infinite sum Σ_(n=1) ^∞ (1/((2n−1)(2n+1)(2n+3)))

$${Find}\:{the}\:{infinite}\:{sum} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$

Commented by mathmax by abdo last updated on 26/Apr/20

let decompose F(x)=(1/((2x−1)(2x+1)(2x+3))) F(x)=(a/(2x−1)) +(b/(2x+1)) +(c/(2x+3)) a =(2x−1)F(x)∣_(x=(1/2)) =(1/((2)(4))) =(1/8) b =(2x+1)F(x)∣_(x=−(1/2)) =(1/((−2)(2)))=−(1/4) c =(2x+3)F(x)∣_(x=−(3/2)) =(1/((−4)(−2))) =(1/8) ⇒ F(x)=(1/(8(2x−1)))−(1/(4(2x+1)))+(1/(8(2x+3))) let S_n =Σ_(k=1) ^(n ) (1/((2k−1)(2k+1)(2k+3))) ⇒S_n =ΣF(k) =(1/8)Σ_(k=1) ^n (1/(2k−1)) −(1/4)Σ_(k=1) ^n (1/(2k+1)) +(1/8)Σ_(k=1) ^n (1/(2k+3)) we have Σ_(k=1) ^n (1/(2k−1)) =1 +(1/3)+(1/5)+....+(1/(2n−1)) =1+(1/2)+(1/3)+...+(1/(2n−1))+(1/(2n))−(1/2)−(1/4)−....−(1/(2n)) =H_(2n) −(1/2)H_n Σ_(k=1) ^n (1/(2k+1)) =_(k=j−1) Σ_(j=2) ^(n+1) (1/(2j−1)) =Σ_(j=1) ^(n+1) (1/(2j−1))−1 =H_(2n+2) −(1/2)H_(n+1) −1 Σ_(k=1) ^n (1/(2k+3)) =_(k=j−2) Σ_(j=3) ^(n+2) (1/(2j−1)) =Σ_(j=1) ^(n+2) (1/(2j−1))−1−(1/3) =H_(2n+4) −(1/2)H_(n+2) −(4/3) ⇒ 8S_n =H_(2n) −(1/2)H_n −2(H_(2n+2) −(1/2)H_(n+1) −1)+H_(2n+4) −(1/2)H_(n+2) −(4/3) =H_(2n) −(1/2)H_n −2H_(2n+2) +H_(n+1) +2 +H_(2n+4) −(1/2)H_(n+2) −(4/3) ∼ln(2n)+γ −(1/2)ln(n)−(γ/2) −2ln(2n+2)−2γ +ln(n+1)+γ +ln(2n+4)+γ −(1/2)ln(n+2)−(γ/2) +(2/3) ⇒ 8S_n ∼ ln(2)+ln(n)−(1/2)ln(n)−2ln(2)−2ln(n)+ln(n) +ln(2)+ln(n)−(1/2)ln(n) +(2/3) ⇒8S_n ∼(2/3) ⇒S_n ∼(1/(12)) ⇒lim_(n→+∞) S_n =(1/(12))

$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{3}\right)} \\ $$$${F}\left({x}\right)=\frac{{a}}{\mathrm{2}{x}−\mathrm{1}}\:+\frac{{b}}{\mathrm{2}{x}+\mathrm{1}}\:+\frac{{c}}{\mathrm{2}{x}+\mathrm{3}} \\ $$$${a}\:=\left(\mathrm{2}{x}−\mathrm{1}\right){F}\left({x}\right)\mid_{{x}=\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:=\frac{\mathrm{1}}{\left(\mathrm{2}\right)\left(\mathrm{4}\right)}\:=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${b}\:=\left(\mathrm{2}{x}+\mathrm{1}\right){F}\left({x}\right)\mid_{{x}=−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\:=\frac{\mathrm{1}}{\left(−\mathrm{2}\right)\left(\mathrm{2}\right)}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${c}\:=\left(\mathrm{2}{x}+\mathrm{3}\right){F}\left({x}\right)\mid_{{x}=−\frac{\mathrm{3}}{\mathrm{2}}} \:\:\:=\frac{\mathrm{1}}{\left(−\mathrm{4}\right)\left(−\mathrm{2}\right)}\:=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{x}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{2}{x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{2}{x}+\mathrm{3}\right)}\:{let} \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}\:} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{3}\right)}\:\Rightarrow{S}_{{n}} =\Sigma{F}\left({k}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{8}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\:\:{we}\:{have} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:=\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+....+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−....−\frac{\mathrm{1}}{\mathrm{2}{n}} \\ $$$$={H}_{\mathrm{2}{n}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:=_{{k}={j}−\mathrm{1}} \:\:\:\sum_{{j}=\mathrm{2}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}{j}−\mathrm{1}}\:=\sum_{{j}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}{j}−\mathrm{1}}−\mathrm{1} \\ $$$$={H}_{\mathrm{2}{n}+\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}+\mathrm{1}} −\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\:=_{{k}={j}−\mathrm{2}} \:\:\:\sum_{{j}=\mathrm{3}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{2}{j}−\mathrm{1}}\:=\sum_{{j}=\mathrm{1}} ^{{n}+\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{2}{j}−\mathrm{1}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$={H}_{\mathrm{2}{n}+\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}+\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{8}{S}_{{n}} ={H}_{\mathrm{2}{n}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} −\mathrm{2}\left({H}_{\mathrm{2}{n}+\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}+\mathrm{1}} −\mathrm{1}\right)+{H}_{\mathrm{2}{n}+\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}+\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$={H}_{\mathrm{2}{n}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}} −\mathrm{2}{H}_{\mathrm{2}{n}+\mathrm{2}} \:+{H}_{{n}+\mathrm{1}} +\mathrm{2}\:+{H}_{\mathrm{2}{n}+\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}{H}_{{n}+\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\sim{ln}\left(\mathrm{2}{n}\right)+\gamma\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)−\frac{\gamma}{\mathrm{2}}\:−\mathrm{2}{ln}\left(\mathrm{2}{n}+\mathrm{2}\right)−\mathrm{2}\gamma\:+{ln}\left({n}+\mathrm{1}\right)+\gamma \\ $$$$+{ln}\left(\mathrm{2}{n}+\mathrm{4}\right)+\gamma\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}+\mathrm{2}\right)−\frac{\gamma}{\mathrm{2}}\:\:+\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{8}{S}_{{n}} \sim\:\:{ln}\left(\mathrm{2}\right)+{ln}\left({n}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{2}{ln}\left({n}\right)+{ln}\left({n}\right) \\ $$$$+{ln}\left(\mathrm{2}\right)+{ln}\left({n}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({n}\right)\:+\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\mathrm{8}{S}_{{n}} \sim\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow{S}_{{n}} \sim\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented by Maclaurin Stickker last updated on 26/Apr/20

Thank you, I appreciate your work.

$${Thank}\:{you},\:{I}\:{appreciate}\:{your}\:{work}. \\ $$

Commented by abdomathmax last updated on 27/Apr/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

Answered by ajfour last updated on 26/Apr/20

S=(1/4)Σ(((2n+3)−(2n−1))/((2n−1)(2n+1)(2n+3))) =(1/4)Σ(1/((2n−1)(2n+1)))−(1/4)Σ(1/((2n+1)(2n+3))) =(1/8)Σ(1/(2n−1))−(1/8)Σ(1/(2n+1))−(1/8)Σ(1/(2n+1))+(1/8)Σ(1/(2n+3)) 8S=((1/1)+(1/3)+(1/5)+(1/7)+...)−2((1/3)+(1/5)+(1/7)+...)+((1/5)+(1/7)+...) =1−(1/3) =(2/3) ⇒ S=(1/(12)) . (any chance this may be correct?)

$${S}=\frac{\mathrm{1}}{\mathrm{4}}\Sigma\frac{\left(\mathrm{2}{n}+\mathrm{3}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{4}}\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}}\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{8}}\Sigma\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{8}}\Sigma\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{8}}\Sigma\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}}\Sigma\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}} \\ $$$$\mathrm{8}{S}=\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{7}}+...\right)−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{7}}+...\right)+\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{7}}+...\right) \\ $$$$\:\:\:\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:{S}=\frac{\mathrm{1}}{\mathrm{12}}\:. \\ $$$$\left({any}\:{chance}\:{this}\:{may}\:{be}\:{correct}?\right) \\ $$

Commented by mathmax by abdo last updated on 26/Apr/20

correct sir ajfour

$${correct}\:{sir}\:{ajfour} \\ $$

Commented by Maclaurin Stickker last updated on 26/Apr/20

Perfect!

$${Perfect}! \\ $$

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