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Question Number 83144 by M±th+et£s last updated on 28/Feb/20

$$showthat$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n-1)(3n-1)}=\frac{1}{6}(\sqrt{3}\pi -9log\left(3\right)+log\left(4096\right))$$

Answered by mind is power last updated on 28/Feb/20

$$=\frac{1}{6}\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{(n-\frac{1}{2})(n-\frac{1}{3})}$$$$=\frac{1}{6}\underset{n=0}{\sum ^{+\mathrm{\infty }}}\frac{1}{(n+\frac{1}{2})(n+\frac{2}{3})}=\frac{1}{6}.\frac{\mathrm{\Psi }\left(\frac{2}{3}\right)-\mathrm{\Psi }\left(\frac{1}{2}\right)}{\frac{2}{3}-\frac{1}{2}}$$$$\mathrm{\Psi }\left(\frac{2}{3}\right)=-\gamma -ln\left(6\right)-\frac{\pi }{2}cot\left(\frac{2\pi }{3}\right)+2cos\left(\frac{4\pi }{3}\right)ln\left(sin\frac{\pi }{3}\right)$$$$\mathrm{\Psi }\left(\frac{1}{2}\right)=-\gamma -ln\left(4\right)$$$$\mathrm{\Psi }\left(\frac{2}{3}\right)-\mathrm{\Psi }\left(\frac{1}{2}\right)=ln\left(\frac{2}{3}\right)+\frac{\pi }{2\sqrt{3}}-ln\left(\frac{\sqrt{3}}{2}\right)$$$$\frac{\mathrm{\Psi }\left(\frac{2}{3}\right)-\mathrm{\Psi }\left(\frac{1}{2}\right)}{\frac{2}{3}-\frac{1}{2}}=6(\mathrm{\Psi }\left(\frac{2}{3}\right)-\mathrm{\Psi }\left(\frac{1}{2}\right))=6(\frac{\pi }{2\sqrt{3}}+ln\left(2\right)-ln\left(3\right)-\frac{1}{2}ln\left(3\right)+ln\left(2\right))$$$$=\pi \sqrt{3}+ln\left(2^{12}\right)-3ln\left(3\right)=\pi \sqrt{3}+ln\left(4096\right)-9ln\left(3\right)$$$$\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{(2n-1)(3n-1)}=\frac{1}{6}.(\pi \sqrt{3}-9ln\left(3\right)+ln\left(4096\right))$$$$$$

Commented by mind is power last updated on 28/Feb/20

$$\mathrm{\Psi }\left(\frac{p}{q}\right)=-\gamma -ln\left(2q\right)-\frac{\pi }{2}cot\left(\frac{p\pi }{q}\right)+2\underset{n=1}{\sum ^{\left[\frac{q-1}{2}\right]}}cos\left(\frac{2\pi np}{q}\right)ln\left(sin\frac{\pi n}{q}\right)$$$$$$

Commented by M±th+et£s last updated on 28/Feb/20

$$godblessyousir$$

Commented by msup trace by abdo last updated on 28/Feb/20

$$define\mathrm{\Phi }sirmind....$$

Commented by mathmax by abdo last updated on 28/Feb/20

$$thankyousirmind$$

Commented by mind is power last updated on 28/Feb/20

$$withepleasursir$$

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