Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 147877 by Tawa11 last updated on 24/Jul/21

Find the sum to close form. (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ...

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{close}\:\mathrm{form}. \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:\:+\:\:\:...\:\:\: \\ $$

Commented by Tinku Tara last updated on 24/Jul/21

Welcome back. Tawa. 🙂🙂

$$\mathrm{Welcome}\:\mathrm{back}.\:\mathrm{Tawa}. \\ $$🙂🙂

Commented by Tawa11 last updated on 24/Jul/21

Thank you sir. (⌣^(⌢ . . ⌢) )

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\:\:\:\:\:\:\:\:\left(\overset{\frown\:\:\:.\:\:\:.\:\:\:\frown} {\smile}\right) \\ $$

Answered by Olaf_Thorendsen last updated on 24/Jul/21

S = Σ_(n=0) ^∞ (1/((3n+1)(3n+2)(3n+3))) S = (1/8)Σ_(n=0) ^∞ (3/((3n+1)(3n+2)))−(5/((3n+2)(3n+3)))+(2/((3n+3)(3n+1))) S = (1/(24))Σ_(n=0) ^∞ (1/((n+(1/3))(n+(2/3))))−(5/(72))Σ_(n=0) ^∞ (1/((n+(2/3))(n+1))) +(1/(36))Σ_(n=0) ^∞ (1/((n+1)(n+(1/3)))) S = (1/(24)).((ψ((2/3))−ψ((1/3)))/((2/3)−(1/3)))−(5/(72)).((ψ(1)−ψ((2/3)))/(1−(2/3))) +(1/(36)).((ψ(1)−ψ((1/3)))/(1−(1/3))) S = (1/8)(ψ((2/3))−ψ((1/3)))−(5/(24))(ψ(1)−ψ((2/3))) +(1/(24))(ψ(1)−ψ((1/3))) S = −(1/6)ψ((1/3))+(1/3)ψ((2/3))−(1/6)ψ(1) • ψ(1−z) = ψ(z)+πcot(πz) ⇒ ψ((2/3)) = ψ((1/3))+πcot((π/3)) ψ((2/3)) = ψ((1/3))+(π/( (√3))) S = (1/6)ψ((1/3))+(π/(3(√3)))−(1/6)ψ(1) • ψ(1) = −γ • ψ((1/3)) = −(π/(2(√3)))−(3/2)ln3−γ S = (1/6)(−(π/(2(√3)))−(3/2)ln3−γ)+(π/(3(√3)))+(γ/6) S = (π/( 4(√3)))−(1/4)ln3

$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{3}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}−\frac{\mathrm{5}}{\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{3}\right)}+\frac{\mathrm{2}}{\left(\mathrm{3}{n}+\mathrm{3}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{24}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)}−\frac{\mathrm{5}}{\mathrm{72}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)\left({n}+\mathrm{1}\right)} \\ $$$$+\frac{\mathrm{1}}{\mathrm{36}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{24}}.\frac{\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{5}}{\mathrm{72}}.\frac{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{36}}.\frac{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\left(\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right)−\frac{\mathrm{5}}{\mathrm{24}}\left(\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{24}}\left(\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$$\mathrm{S}\:=\:−\frac{\mathrm{1}}{\mathrm{6}}\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{3}}\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{6}}\psi\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\bullet\:\psi\left(\mathrm{1}−{z}\right)\:=\:\psi\left({z}\right)+\pi\mathrm{cot}\left(\pi{z}\right) \\ $$$$\Rightarrow\:\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:=\:\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\pi\mathrm{cot}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\:=\:\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{6}}\psi\left(\mathrm{1}\right) \\ $$$$\bullet\:\psi\left(\mathrm{1}\right)\:=\:−\gamma \\ $$$$\bullet\:\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln3}−\gamma \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left(−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln3}−\gamma\right)+\frac{\pi}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{\gamma}{\mathrm{6}} \\ $$$$\mathrm{S}\:=\:\frac{\pi}{\:\mathrm{4}\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln3} \\ $$

Commented by Tawa11 last updated on 24/Jul/21

Wow, God bless you sir.

$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 24/Jul/21

Sir, where can I learn summation like this. Please sir prescribe book. I want to study on it too. Thanks sir.

$$\mathrm{Sir},\:\mathrm{where}\:\mathrm{can}\:\mathrm{I}\:\mathrm{learn}\:\mathrm{summation}\:\mathrm{like}\:\mathrm{this}.\:\mathrm{Please}\:\mathrm{sir}\:\mathrm{prescribe}\:\mathrm{book}.\:\mathrm{I}\:\mathrm{want} \\ $$$$\mathrm{to}\:\mathrm{study}\:\mathrm{on}\:\mathrm{it}\:\mathrm{too}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 24/Jul/21

Sir. Can we find sum to n terms here: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ...

$$\mathrm{Sir}. \\ $$$$\mathrm{Can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{sum}\:\mathrm{to}\:\:\mathrm{n}\:\:\mathrm{terms}\:\mathrm{here}:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:+\:... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com