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Question Number 167517 by Gbenga last updated on 18/Mar/22

$$\underset{\mathbf{n}=1}{\mathbf{\sum }^{\mathrm{\infty }}}\frac{\mathbf{cos}\left(\mathbf{n}\right)\mathbf{sin}\left(\mathbf{n}\right)}{\mathbf{tan}\left(\mathbf{n}\right)}$$

Answered by alephzero last updated on 18/Mar/22

$$\frac{\cos n\sin n}{\tan n}=\frac{\cos n\sin n}{\frac{\sin n}{\cos n}}=$$$$=\frac{\cos n\sin n\cos n}{\sin n}={\cos }^{2}n$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\cos n\sin n}{\tan n}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{\cos }^{2}n$$$$\rho =\underset{n\to \mathrm{\infty }}{\lim }\mid \frac{a_{n+1}}{a_n}\mid =\lim \frac{{\cos }^2(n+1)}{{\cos }^{2}n}=$$$$={\left(\lim \frac{\cos (n+1)}{\cos n}\right)}^2\Rightarrow \mathrm{indeterminate}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\cos n\sin n}{\tan n}\mathrm{diverges}\mathrm{to}+\mathrm{\infty }$$

Commented by Gbenga last updated on 18/Mar/22

$$\mathbf{thanks}\mathbf{sir}$$

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