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Question Number 31706 by gunawan last updated on 13/Mar/18

$$\mathrm{Given}{\theta }_n=arc\tan n,\underset{x\to \mathrm{\infty }}{\lim }{\theta }_{n+1}-{\theta }_n=$$

Commented by abdo imad last updated on 18/Mar/18

$$ifthequestionisfind{lim}_{n\to \mathrm{\infty }}{\theta }_{n+1}-{\theta }_{n}wehave$$$$lim{\theta }_{n+1}-{\theta }_n={lim}_{n\to \mathrm{\infty }}arctan(n+1)-arctann$$$$\frac{\pi }{2}-\frac{\pi }{2}=0$$

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