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 51985 by maxmathsup by imad last updated on 01/Jan/19

1) let p integr natural not 0 calculate arctan((p/(p+1)))−arctan(((p−1)/p))  2)let S_n =Σ_(p=1) ^n  arctan((1/(2p^2 ))) find lim_(n→+∞)  S_n

$$\left.\mathrm{1}\right)\:{let}\:{p}\:{integr}\:{natural}\:{not}\:\mathrm{0}\:{calculate}\:{arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right) \\ $$$$\left.\mathrm{2}\right){let}\:{S}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \\ $$

Commented by Abdo msup. last updated on 19/Jan/19

we havetan( arctan((p/(p+1)))−arctan(((p−1)/p)))  =(((p/(p+1))−((p−1)/p))/(1+(p/(p+1))((p−1)/p))) =((p^2 −p^2 +1)/(p^2 +p+p^2 −p)) =(1/(2p^2 )) .  2) we have arctan((1/(2p^2 )))=arctan((p/(p+1)))−arctan(((p−1)/p))  we haveS_n =Σ_(p=1) ^n  arctan((1/(2p^2 ))) ⇒  S_n =Σ_(p=1) ^n (arctan((p/(p+1)))−arctan(((p−1)/p)))  =Σ_(p=1) ^n  (U_p −U_(p−1) )         (U_p =arctan((p/(p+1))))  =U_1 −U_0 +U_2 −U_1 +....+U_n −U_(n−1)   =U_n −U_1 =arctan((n/(n+1)))−arctan((1/2)) ⇒  lim_(n→+∞)  S_n =(π/4) −arctan((1/2)).

$${we}\:{havetan}\left(\:{arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right)\right) \\ $$$$=\frac{\frac{{p}}{{p}+\mathrm{1}}−\frac{{p}−\mathrm{1}}{{p}}}{\mathrm{1}+\frac{{p}}{{p}+\mathrm{1}}\frac{{p}−\mathrm{1}}{{p}}}\:=\frac{{p}^{\mathrm{2}} −{p}^{\mathrm{2}} +\mathrm{1}}{{p}^{\mathrm{2}} +{p}+{p}^{\mathrm{2}} −{p}}\:=\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\:. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)={arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right) \\ $$$${we}\:{haveS}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${S}_{{n}} =\sum_{{p}=\mathrm{1}} ^{{n}} \left({arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)−{arctan}\left(\frac{{p}−\mathrm{1}}{{p}}\right)\right) \\ $$$$=\sum_{{p}=\mathrm{1}} ^{{n}} \:\left({U}_{{p}} −{U}_{{p}−\mathrm{1}} \right)\:\:\:\:\:\:\:\:\:\left({U}_{{p}} ={arctan}\left(\frac{{p}}{{p}+\mathrm{1}}\right)\right) \\ $$$$={U}_{\mathrm{1}} −{U}_{\mathrm{0}} +{U}_{\mathrm{2}} −{U}_{\mathrm{1}} +....+{U}_{{n}} −{U}_{{n}−\mathrm{1}} \\ $$$$={U}_{{n}} −{U}_{\mathrm{1}} ={arctan}\left(\frac{{n}}{{n}+\mathrm{1}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\pi}{\mathrm{4}}\:−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right). \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com