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Question Number 140405 by mnjuly1970 last updated on 07/May/21

$$$$$$findthevalueof::$$$$$$$$\mathrm{\Theta }:=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4n.(4n+1).(4n+2).(4n+3)}=?$$$$$$

Answered by qaz last updated on 07/May/21

$$f\left(n\right)=\frac{1}{4n(4n+1)(4n+2)(4n+3)}=\frac{A}{4n}+\frac{B}{4n+1}+\frac{C}{4n+2}+\frac{D}{4n+3}$$$$A=\underset{4n\to 0}{\lim 4}nf\left(n\right)=\frac{1}{6}B=\underset{4n\to -1}{\lim }(4n+1)f\left(n\right)=-\frac{1}{2}$$$$C=\underset{4n\to -2}{\lim }(4n+2)f\left(n\right)=\frac{1}{2}D=\underset{4n\to -3}{\lim }(4n+3)f\left(n\right)=-\frac{1}{6}$$$$\mathrm{\Theta }=\underset{n=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{6\times 4n}-\frac{1}{2(4n+1)}+\frac{1}{2(4n+2)}-\frac{1}{6(4n+3)}]$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}[\frac{1}{24n}-\frac{1}{8(n+\frac{1}{4})}+\frac{1}{8(n+\frac{1}{2})}-\frac{1}{24(n+\frac{3}{4})}]$$$$=\frac{1}{24}\psi \left(\frac{7}{4}\right)+\frac{1}{8}\psi \left(\frac{5}{4}\right)-\frac{1}{8}\psi \left(\frac{3}{2}\right)$$$$=-\frac{1}{24}\gamma +\frac{11}{36}-\frac{\pi }{24}-\frac{1}{4}ln2$$

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