Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 33716 by prof Abdo imad last updated on 22/Apr/18

$$calculate\sum _{n=0}^{\mathrm{\infty }}arctan\left(\frac{e^{n+1}-e^n}{1+e^{2n+1}}\right).$$

Commented by prof Abdo imad last updated on 31/May/18

$$letput{S}_n=\sum _{k=0}^{n}arctan\left(\frac{e^{n+1}-e^n}{1+e^{2n+1}}\right)letuse$$$$thechangement{e}^n=tan\left(u_n\right)\Rightarrow u_n=arctan\left(e^n\right)$$$$\frac{e^{n+1}-e^n}{1+e^{2n+1}}=\frac{tan\left(u_{n+1}\right)-tan\left(u_n\right)}{1+tan\left(u_{n+1}\right)tan\left(u_n\right)}$$$$=tan(u_{n+1}-u_n)\Rightarrow S_n=\sum _{k=0}^n(u_{n+1}-u_n)$$$$=u_{n+1}-u_0=arctan\left(e^{n+1}\right)-\frac{\pi }{4}$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$$$$★{lim}_{n\to +\mathrm{\infty }}{S}_n=\frac{\pi }{4}★$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18

$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\{{\tan }^{-1}\left(e^{n+1}\right)-{\tan }^{-1}\left(e^n\right)\}$$$$T_n={\tan }^{-1}\left(e^{n+1}\right)-{\tan }^{-1}\left(e^n\right)$$$$T_o={\tan }^{-1}\left(e^1\right)-{\tan }^{-1}\left(e^0\right)$$$$T_1={\tan }^{-1}\left(e^2\right)-{\tan }^{-1}\left(e^1\right)$$$$T_2={\tan }^{-1}\left(e^3\right)-{\tan }^{-1}\left(e^2\right)$$$$soaddinguptonterms$$$$S_n={\tan }^{-1}\left(e^{n+1}\right)-{\tan }^{-1}\left(e^o\right)$$$$whenntendstoinfinity$$$$={\tan }^{-1}\left(\mathrm{\infty }\right)-{\tan }^{-1}\left(1\right)$$$$=\mathrm{\Pi }/2-\mathrm{\Pi }/4$$$$=\mathrm{\Pi }/4$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com