Question-151573

Question Number 151573 by amin96 last updated on 22/Aug/21

Answered by Olaf_Thorendsen last updated on 22/Aug/21

S_n = Σ_(k=0) ^(n−1) (1/((3k+1)(3k+2)(3k+3))) S_n = Σ_(k=0) ^(n−1) (((1/2)/(3k+1))−(1/(3k+2))+((1/2)/(3k+3))) S_n = (1/2)Σ_(k=0) ^(n−1) (1/(3k+1))−Σ_(k=0) ^(n−1) (1/(3k+2))+(H_n /6) S_n = (1/2)(((Ψ(n+(1/3)))/3)+(γ/3)+((π(√3))/(18))+((ln3)/2)) −(((Ψ(n+(2/3)))/3)+(γ/3)−((π(√3))/(18))+((ln3)/2))+(H_n /6) S_n = ((Ψ(n+(1/3)))/6)−((Ψ(n+(2/3)))/3)−(γ/6)+((π(√3))/(12))−((ln3)/4)+(H_n /6)

$$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{k}+\mathrm{1}\right)\left(\mathrm{3}{k}+\mathrm{2}\right)\left(\mathrm{3}{k}+\mathrm{3}\right)} \\ $$$$\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{\mathrm{1}/\mathrm{2}}{\mathrm{3}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{2}}+\frac{\mathrm{1}/\mathrm{2}}{\mathrm{3}{k}+\mathrm{3}}\right) \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}−\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{2}}+\frac{{H}_{{n}} }{\mathrm{6}} \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\Psi\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{3}}+\frac{\gamma}{\mathrm{3}}+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}+\frac{\mathrm{ln3}}{\mathrm{2}}\right) \\ $$$$−\left(\frac{\Psi\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{3}}+\frac{\gamma}{\mathrm{3}}−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}+\frac{\mathrm{ln3}}{\mathrm{2}}\right)+\frac{{H}_{{n}} }{\mathrm{6}} \\ $$$$\mathrm{S}_{{n}} \:=\:\frac{\Psi\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{6}}−\frac{\Psi\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{3}}−\frac{\gamma}{\mathrm{6}}+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{12}}−\frac{\mathrm{ln3}}{\mathrm{4}}+\frac{{H}_{{n}} }{\mathrm{6}} \\ $$

Commented by Tawa11 last updated on 22/Aug/21

$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 22/Aug/21

Please one more sir. How is: S_n = Σ_(k = 0) ^(n − 1) (1/(3k + 1)) = ((ψ(n + (1/3)))/3) + (γ/3) + ((π(√3))/(18)) + ((ln 3)/2) Thanks sir. Help me to understand.

$$ \\ $$$$\mathrm{Please}\:\mathrm{one}\:\mathrm{more}\:\mathrm{sir}. \\ $$$$\:\:\:\mathrm{How}\:\mathrm{is}:\:\:\:\:\:\:\:\:\mathrm{S}_{\mathrm{n}} \:\:\:=\:\:\:\underset{\mathrm{k}\:\:=\:\:\mathrm{0}} {\overset{\mathrm{n}\:\:\:−\:\:\:\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3k}\:\:+\:\:\mathrm{1}}\:\:\:\:\:=\:\:\:\:\frac{\psi\left(\mathrm{n}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{3}}\right)}{\mathrm{3}}\:\:\:+\:\:\:\frac{\gamma}{\mathrm{3}}\:\:\:+\:\:\:\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{18}}\:\:\:+\:\:\:\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{Help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{understand}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply