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 148946 by Tawa11 last updated on 01/Aug/21

Sum to n term: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ... to n.

$$\mathrm{Sum}\:\mathrm{to}\:\:\mathrm{n}\:\:\mathrm{term}:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}.\mathrm{5}.\mathrm{6}}\:\:\:+\:\:\:\frac{\mathrm{1}}{\mathrm{7}.\mathrm{8}.\mathrm{9}}\:\:+\:\:...\:\:\:\mathrm{to}\:\:\mathrm{n}. \\ $$

Answered by Olaf_Thorendsen last updated on 02/Aug/21

S = Σ_(n=0) ^∞ (1/((3n+1)(3n+2)(3n+3))) S = Σ_(n=0) ^∞ ((1/((3n+1)(3n+2)))−(1/((3n+1)(3n+3)))) S = (1/9)Σ_(n=0) ^∞ ((1/((n+(1/3))(n+(2/3))))−(1/((n+(1/3))(n+1)))) S = (1/9)Σ_(n=0) ^∞ (((ψ((2/3))−ψ((1/3)))/((2/3)−(1/3)))−((ψ(1)−ψ((1/3)))/(1−(1/3)))) • ψ(1) = −γ • ψ((1/3)) = −(π/(2(√3)))−(3/2)ln3−γ • ψ(1−z) −ψ(z) = πcot(πz) ⇒ ψ(1−(1/3))−ψ((1/3)) = πcot((π/3)) ψ((2/3))−ψ((1/3)) = (π/( (√3))) S = (1/9)(((π/( (√3)))/(1/3))−((−γ+(π/(2(√3)))+(3/2)ln3+γ)/(2/3))) S = (1/9)(π(√3)−((π(√3))/( 4))−(9/4)ln3) S = (π/( 4(√3)))−(1/4)ln3

$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}−\frac{\mathrm{1}}{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{3}\right)}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)}−\frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left({n}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\psi\left(\mathrm{1}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}}\right) \\ $$$$\bullet\:\psi\left(\mathrm{1}\right)\:=\:−\gamma \\ $$$$\bullet\:\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln3}−\gamma \\ $$$$\bullet\:\psi\left(\mathrm{1}−{z}\right)\:−\psi\left({z}\right)\:=\:\pi\mathrm{cot}\left(\pi{z}\right) \\ $$$$\Rightarrow\:\psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:\pi\mathrm{cot}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\:=\:\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\frac{\pi}{\:\sqrt{\mathrm{3}}}}{\frac{\mathrm{1}}{\mathrm{3}}}−\frac{−\gamma+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{ln3}+\gamma}{\frac{\mathrm{2}}{\mathrm{3}}}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\left(\pi\sqrt{\mathrm{3}}−\frac{\pi\sqrt{\mathrm{3}}}{\:\mathrm{4}}−\frac{\mathrm{9}}{\mathrm{4}}\mathrm{ln3}\right) \\ $$$$\mathrm{S}\:=\:\frac{\pi}{\:\mathrm{4}\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln3} \\ $$

Commented by Tawa11 last updated on 02/Aug/21

Thanks sir. God bless you.

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com