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Question Number 155420 by amin96 last updated on 30/Sep/21

$$\frac{1}{42}+\frac{1}{72}+\frac{1}{110}+\dots =?$$

Answered by puissant last updated on 01/Oct/21

$$S=\frac{1}{42}+\frac{1}{72}+\frac{1}{110}+.......=\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+....$$$$\Rightarrow S=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{(2n+2)(2n+3)}=\frac{1}{4}\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)(n+\frac{3}{2})}$$$$=\frac{1}{4}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(n+1)(n+\frac{3}{2})}-\frac{1}{20}-\frac{1}{6}$$$$=\frac{1}{4}\bullet \frac{\psi \left(\frac{3}{2}\right)-\psi \left(1\right)}{\frac{3}{2}-1}-\frac{13}{60}=\frac{1}{2}(\psi \left(\frac{3}{2}\right)-\psi \left(1\right))-\frac{13}{60}$$$$\psi \left(\frac{3}{2}\right)=\psi (1+\frac{1}{2})=\psi \left(\frac{1}{2}\right)+2;$$$$\psi \left(\frac{1}{2}\right)=-2ln2-\gamma ;\psi \left(1\right)=-\gamma ..$$$$\Rightarrow S=\frac{1}{2}(-2ln2-\gamma +2+\gamma )-\frac{13}{60}$$$$\Rightarrow S=-ln2+1-\frac{13}{60}..$$$$$$$$\therefore \because S=\frac{47}{60}-ln2...$$$$$$$$....................\mathcal{L}epuissant..................$$

Commented by amin96 last updated on 30/Sep/21

$$greatsir$$

Commented by Tawa11 last updated on 30/Sep/21

$$\mathrm{Nice}\mathrm{sir}$$

Commented by puissant last updated on 30/Sep/21

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Commented by SANOGO last updated on 30/Sep/21

$$lepuissant$$

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