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n-0-tan-1-n-1-2-tan-1-n-1-2-




Question Number 30929 by rahul 19 last updated on 28/Feb/18
Σ_(n=0) ^∞  tan^(−1) (n+(1/2))−tan^(−1) (n−(1/2))= ?
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\mathrm{tan}^{−\mathrm{1}} \left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{tan}^{−\mathrm{1}} \left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\:? \\ $$
Commented by abdo imad last updated on 01/Mar/18
let put S_n = Σ_(k=0) ^n  arctan(k+(1/2))−arctan(k−(1/2))  S_n =Σ_(k=0) ^n arctan(((2k+1)/2))−actan(((2k−1)/2))  =Σ_(k=0) ^n  (v_k  −v_(k−1) ) with v_k =arctan(((2k+1)/2))  S_n =v_0  −v_(−1)  +v_1  −v_o  +...+v_n  −v_(n−1) =v_n  −v_(−1)  ⇒  S_n =arctan(((2n+1)/2)) −arctan(−(1/2)) ⇒  lim_(n→∞)  S_n = (π/2) +arctan((1/2))=(π/2) +(π/2) −arctan2  =π −arctan2 .
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{arctan}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)−{arctan}\left({k}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} {arctan}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}\right)−{actan}\left(\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left({v}_{{k}} \:−{v}_{{k}−\mathrm{1}} \right)\:{with}\:{v}_{{k}} ={arctan}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}\right) \\ $$$${S}_{{n}} ={v}_{\mathrm{0}} \:−{v}_{−\mathrm{1}} \:+{v}_{\mathrm{1}} \:−{v}_{{o}} \:+…+{v}_{{n}} \:−{v}_{{n}−\mathrm{1}} ={v}_{{n}} \:−{v}_{−\mathrm{1}} \:\Rightarrow \\ $$$${S}_{{n}} ={arctan}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right)\:−{arctan}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \:{S}_{{n}} =\:\frac{\pi}{\mathrm{2}}\:+{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\:−{arctan}\mathrm{2} \\ $$$$=\pi\:−{arctan}\mathrm{2}\:. \\ $$
Answered by sma3l2996 last updated on 28/Feb/18
A=Σ_(n=0) ^∞ tan^(−1) (((2n+1)/2))−Σ_(n=0) ^∞ tan^(−1) (((2n−1)/2))  let  k=2n+1  and  i=2n−1  so  A=Σ_(k=1) ^∞ tan^(−1) (k/2)−Σ_(i=−1) ^∞ tan^(−1) (i/2)  =Σ_(k=1) ^∞ tan^(−1) (k/2)−Σ_(i=1) ^∞ tan^(−1) (i/2)−tan^(−1) (−1/2)−tan^(−1) (0)  =−(−π)=π
$${A}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${let}\:\:{k}=\mathrm{2}{n}+\mathrm{1}\:\:{and}\:\:{i}=\mathrm{2}{n}−\mathrm{1} \\ $$$${so}\:\:{A}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left({k}/\mathrm{2}\right)−\underset{{i}=−\mathrm{1}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left({i}/\mathrm{2}\right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left({k}/\mathrm{2}\right)−\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left({i}/\mathrm{2}\right)−{tan}^{−\mathrm{1}} \left(−\mathrm{1}/\mathrm{2}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{0}\right) \\ $$$$=−\left(−\pi\right)=\pi \\ $$
Answered by ajfour last updated on 28/Feb/18
= tan^(−1) ((1/2))−tan^(−1) (((−1)/2))+       tan^(−1) ((3/2))−tan^(−1) ((1/2))+      ...........................+      ...........................+        +tan^(−1) (N+(1/2))−tan^(−1) (N−(1/2))    =  tan^(−1) (N+(1/2))−tan^(−1) (((−1)/2))   with N→ ∞    = (π/2)+tan^(−1) ((1/2))  .
$$=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\mathrm{2}}\right)+ \\ $$$$\:\:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)+ \\ $$$$\:\:\:\:………………………+ \\ $$$$\:\:\:\:………………………+ \\ $$$$\:\:\:\:\:\:+\mathrm{tan}^{−\mathrm{1}} \left({N}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{tan}^{−\mathrm{1}} \left({N}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:=\:\:\mathrm{tan}^{−\mathrm{1}} \left({N}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:{with}\:{N}\rightarrow\:\infty \\ $$$$\:\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:. \\ $$

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