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Sum-to-n-term-1-1-2-3-1-4-5-6-1-7-8-9-to-n-

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

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