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Question Number 73818 by rajesh4661kumar@gmail.com last updated on 16/Nov/19

Commented by Tinku Tara last updated on 16/Nov/19

$$\mathrm{Can}\mathrm{you}\mathrm{post}\mathrm{a}\mathrm{clear}\mathrm{image}.$$

Answered by Tanmay chaudhury last updated on 16/Nov/19

$$T_r={tan}^{-1}\left(\frac{1}{1+r+r^2}\right)={tan}^{-1}\left[\frac{(r+1)-r}{1+r(r+1)}\right]={tan}^{-1}(r+1)-{tan}^{-1}\left(r\right)$$$$S_n=\underset{r=1}{\sum ^n}{T}_r$$$$T_1={tan}^{-1}2-{tan}^{-1}1$$$$T_2={tan}^{-1}3-{tan}^{-1}2$$$$...$$$$...$$$$T_n={tan}^{-1}(n+1)-{tan}^{-1}n$$$$addthem$$$$S_n={tan}^{-1}(n+1)-{tan}^{-1}1$$$$S_{\mathrm{\infty }}={tan}^{-1}\left(\mathrm{\infty }\right)-{tan}^{-1}1$$$$=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}Answer$$

Commented by mind is power last updated on 17/Nov/19

$$\mathrm{nice}\mathrm{worck}\mathrm{sir}$$

Commented by Tanmay chaudhury last updated on 17/Nov/19

$$thankyousir$$

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