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Question Number 163435 by amin96 last updated on 06/Jan/22

Answered by Mathspace last updated on 07/Jan/22

$$S=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{4n+1}(\frac{1}{4n+3}-\frac{1}{4n+4})$$$$=\sum _{m=0}^{\mathrm{\infty }}\frac{1}{(4n+1)(4n+3)}$$$$-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(4n+1)(4n+4)}$$$$=\frac{1}{16}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\frac{1}{4})(n+\frac{3}{4})}$$$$-\frac{1}{16}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\frac{1}{4})(n+1)}$$$$wehave\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+a)(n+b)}$$$$=\frac{\psi \left(a\right)-\psi \left(b\right)}{a-b}\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\frac{1}{4})(n+\frac{3}{4})}$$$$=\frac{\psi \left(\frac{1}{4}\right)-\psi \left(\frac{3}{4}\right)}{\frac{1}{4}-\frac{3}{4}}=\frac{1}{2}(\psi \left(\frac{3}{4}\right)-\psi \left(\frac{1}{4}\right)\}$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\frac{1}{4})(n+1)}$$$$=\frac{\psi \left(\frac{1}{4}\right)-\psi \left(1\right)}{\frac{1}{4}-1}=\frac{4}{3}\{\psi \left(1\right)-\psi \left(\frac{1}{4}\right)\}\Rightarrow$$$$S=\frac{1}{32}\{\psi \left(\frac{3}{4}\right)-\psi \left(\frac{1}{4}\right)\}$$$$-\frac{1}{12}\{\psi \left(1\right)-\psi \left(\frac{1}{4}\right)\}$$$$$$

Commented by Tawa11 last updated on 07/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

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