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Question Number 145082 by mathdanisur last updated on 02/Jul/21

$$\underset{n=2}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{2n^2-2}\right)=?$$

Answered by phally last updated on 02/Jul/21

$$=(-\frac{1}{2})\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\frac{(\mathrm{n}-1)-(\mathrm{n}+1)}{2(\mathrm{n}-1)(\mathrm{n}+1)}$$$$=(-\frac{1}{4})\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}-1})$$$$=(-\frac{1}{4})\underset{x\to \mathrm{\infty }}{\lim }(\frac{1}{\mathrm{n}+1}-1)$$$$=\frac{1}{4}$$

Commented by mathdanisur last updated on 02/Jul/21

$$ThankyouSir,butanswer\frac{3}{8}$$

Commented by phally last updated on 02/Jul/21

$$\mathrm{why}\mathrm{answer}\frac{3}{8}?$$

Commented by mathdanisur last updated on 02/Jul/21

$$yesSerthankyou$$

Answered by Ar Brandon last updated on 02/Jul/21

$$\mathrm{S}\left(\mathrm{n}\right)=\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\frac{1}{{2\mathrm{n}}^2-2}=\frac{1}{2}\sum _{\mathrm{n}⩾2}\frac{1}{{\mathrm{n}}^2-1}$$$$=\frac{1}{2}\sum _{\mathrm{n}⩾2}\frac{1}{(\mathrm{n}-1)(\mathrm{n}+1)}=\frac{1}{2}\sum _{\mathrm{n}⩾2}(\frac{1}{2(\mathrm{n}-1)}-\frac{1}{2(\mathrm{n}+1)})$$$$=\frac{1}{4}[(\frac{1}{1}-\cancel{\frac{1}{3}})+(\frac{1}{2}-\cancel{\frac{1}{4}})+(\cancel{\frac{1}{3}}-\cancel{\frac{1}{5}})+(\cancel{\frac{1}{4}}-\cancel{\frac{1}{6}})+\cdot \cdot \cdot ]$$$$=\frac{1}{4}(1+\frac{1}{2})=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$$

Commented by mathdanisur last updated on 02/Jul/21

$$thankyouSercool$$

Answered by bobhans last updated on 02/Jul/21

$$\mathrm{Find}\mathrm{thd}\mathrm{value}\mathrm{of}\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{{2\mathrm{n}}^2-2}\right).$$$$\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\frac{1}{2({\mathrm{n}}^2-1)}=\frac{1}{2}\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{(\mathrm{n}-1)(\mathrm{n}+1)}\right)$$$$=-\frac{1}{4}\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }\underset{\mathrm{n}=2}{\sum ^{\mathrm{k}}}(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}-1})$$$$=-\frac{1}{4}\underset{\mathrm{k}\to \mathrm{\infty }}{\lim }(\cancel{\frac{1}{3}}-1+\cancel{\frac{1}{4}}-\frac{1}{2}+\cancel{\frac{1}{5}}-\cancel{\frac{1}{3}}+\cancel{\frac{1}{6}}-\cancel{\frac{1}{5}}+...+\frac{1}{\mathrm{k}+1}-\frac{1}{\mathrm{k}-1})$$$$=-\frac{1}{4}.(-1-\frac{1}{2})=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$$

Commented by mathdanisur last updated on 02/Jul/21

$$thankyouSercool$$

Answered by qaz last updated on 02/Jul/21

$$\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{{2\mathrm{n}}^2-2}\right)=\frac{1}{4}\underset{\mathrm{n}=2}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{n}-1}-\frac{1}{\mathrm{n}+1})=\frac{1}{4}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}(\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+3})$$$$=\frac{1}{4}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1({\mathrm{x}}^{\mathrm{n}}-{\mathrm{x}}^{\mathrm{n}+2})\mathrm{dx}=\frac{1}{4}{\int }_0^1(1+\mathrm{x})\mathrm{dx}=\frac{3}{8}$$

Commented by mathdanisur last updated on 02/Jul/21

$$alotcoolSerthankyou..$$

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