Question-175953

Question Number 175953 by rexford last updated on 10/Sep/22

Answered by Ar Brandon last updated on 10/Sep/22

θ=arctan((1/(n^2 +n+1)))=arctan((((n+1)−n)/(1+n(n+1)))) tanθ=(((n+1)−n)/(1+n(n+1)))=((tan(arctan(n+1))−tan(arctan(n)))/(1+tan(arctan(n+1))tan(arctan(n)))) ⇒tanθ=tan((arctan(n+1)−arctan(n)) ⇒θ=arctan(n+1)−arctan(n) ⇒Σ_(n=1) ^m θ_n =Σ_(n=1) ^m (arctan(n+1)−arctan(n)) =arctan(m+1)−arctan(1) =arctan(m+1)−(π/4)

$$\theta=\mathrm{arctan}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right)=\mathrm{arctan}\left(\frac{\left({n}+\mathrm{1}\right)−{n}}{\mathrm{1}+{n}\left({n}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{tan}\theta=\frac{\left({n}+\mathrm{1}\right)−{n}}{\mathrm{1}+{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{tan}\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)\right)−\mathrm{tan}\left(\mathrm{arctan}\left({n}\right)\right)}{\mathrm{1}+\mathrm{tan}\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)\right)\mathrm{tan}\left(\mathrm{arctan}\left({n}\right)\right)} \\ $$$$\Rightarrow\mathrm{tan}\theta=\mathrm{tan}\left(\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)−\mathrm{arctan}\left({n}\right)\right)\right. \\ $$$$\Rightarrow\theta=\mathrm{arctan}\left({n}+\mathrm{1}\right)−\mathrm{arctan}\left({n}\right) \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}\theta_{{n}} =\underset{{n}=\mathrm{1}} {\overset{{m}} {\sum}}\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)−\mathrm{arctan}\left({n}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{arctan}\left({m}+\mathrm{1}\right)−\mathrm{arctan}\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{arctan}\left({m}+\mathrm{1}\right)−\frac{\pi}{\mathrm{4}} \\ $$

Commented by rexford last updated on 10/Sep/22

$${thank}\:{you}\:{very}\:{much} \\ $$

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