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

Answered by Olaf_Thorendsen last updated on 22/Aug/21

$${\mathrm{S}}_n=\underset{k=0}{\sum ^{n-1}}\frac{1}{(3k+1)(3k+2)(3k+3)}$$$${\mathrm{S}}_n=\underset{k=0}{\sum ^{n-1}}(\frac{1/2}{3k+1}-\frac{1}{3k+2}+\frac{1/2}{3k+3})$$$${\mathrm{S}}_n=\frac{1}{2}\underset{k=0}{\sum ^{n-1}}\frac{1}{3k+1}-\underset{k=0}{\sum ^{n-1}}\frac{1}{3k+2}+\frac{H_n}{6}$$$${\mathrm{S}}_n=\frac{1}{2}(\frac{\mathrm{\Psi }(n+\frac{1}{3})}{3}+\frac{\gamma }{3}+\frac{\pi \sqrt{3}}{18}+\frac{\ln 3}{2})$$$$-(\frac{\mathrm{\Psi }(n+\frac{2}{3})}{3}+\frac{\gamma }{3}-\frac{\pi \sqrt{3}}{18}+\frac{\ln 3}{2})+\frac{H_n}{6}$$$${\mathrm{S}}_n=\frac{\mathrm{\Psi }(n+\frac{1}{3})}{6}-\frac{\mathrm{\Psi }(n+\frac{2}{3})}{3}-\frac{\gamma }{6}+\frac{\pi \sqrt{3}}{12}-\frac{\ln 3}{4}+\frac{H_n}{6}$$

Commented by Tawa11 last updated on 22/Aug/21

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

Commented by Tawa11 last updated on 22/Aug/21

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

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