Question Number 51985 by maxmathsup by imad last updated on 01/Jan/19
$$\left.\mathrm{1}\right)\:{let}\:{p}\:{integr}\:{natural}\:{not}\:\mathrm{0}\:{calculate}\:{arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right) \\ $$$$\left.\mathrm{2}\right){let}\:{S}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$
Commented by Abdo msup. last updated on 19/Jan/19
$${we}\:{havetan}\left(\:{arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right)\right) \\ $$$$=\frac{\frac{{p}}{{p}+\mathrm{1}}−\frac{{p}−\mathrm{1}}{{p}}}{\mathrm{1}+\frac{{p}}{{p}+\mathrm{1}}\frac{{p}−\mathrm{1}}{{p}}}\:=\frac{{p}^{\mathrm{2}} −{p}^{\mathrm{2}} +\mathrm{1}}{{p}^{\mathrm{2}} +{p}+{p}^{\mathrm{2}} −{p}}\:=\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\:. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)={arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right) \\ $$$${we}\:{haveS}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${S}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \left({arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right)\right) \\ $$$$=\sum_{{p}=\mathrm{1}} ^{{n}} \:\left({U}_{{p}} −{U}_{{p}−\mathrm{1}} \right)\:\:\:\:\:\:\:\:\:\left({U}_{{p}} ={arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)\right) \\ $$$$={U}_{\mathrm{1}} −{U}_{\mathrm{0}} +{U}_{\mathrm{2}} −{U}_{\mathrm{1}} +….+{U}_{{n}} −{U}_{{n}−\mathrm{1}} \\ $$$$={U}_{{n}} −{U}_{\mathrm{1}} ={arctan}\left(\frac{{n}}{{n}+\mathrm{1}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\pi}{\mathrm{4}}\:−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right). \\ $$