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Question Number 33718 by prof Abdo imad last updated on 22/Apr/18
find Σ_(n=1) ^(+∞) arctan( ((ln(1+(1/(n+1))) −ln(1+(1/n)))/(1+(1 +(1/n))(1+(1/(n+1))))))
$${find}\:\:\sum_{{n}=\mathrm{1}} ^{+\infty} {arctan}\left(\:\frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\:−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}{\mathrm{1}+\left(\mathrm{1}\:+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\left.{n}+\mathrm{1}\right)}\right.}\right) \\ $$
Commented by prof Abdo imad last updated on 26/Apr/18
let put S_n =Σ_(n=1) ^N arctan( ((ln(1+(1/(n+1)))−ln(1+(1/n)))/(1+ln(1+(1/n))ln(1+(1/(n+1)))))) let put ln(1+(1/n))=tan(u_n )⇒u_n =arctan(ln(1+(1/n))) S_n =Σ_(k=1) ^n arctan( ((tan(u_(n+1) ) −tan(u_n ))/(1+tan(u_n )tan(u_(n+1) )))) =Σ_(k=1) ^N arctan(tan(u_(n+1) −u_n )) =Σ_(k=1) ^N (u_(n+1) −u_n )= u_(N+1) −u_1 =arctan(ln(1+(1/(N+1))) −arctan(ln(2)) and lim_(n→+∞) S_n =arctan(ln(1))−arctan(ln(2)) lim_(n→+∞) S_n =−arctan(ln(2)) .
$${let}\:{put}\:{S}_{{n}} =\sum_{{n}=\mathrm{1}} ^{{N}} \:{arctan}\left(\:\frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}{\mathrm{1}+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)}\right) \\ $$$${let}\:{put}\:{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)={tan}\left({u}_{{n}} \right)\Rightarrow{u}_{{n}} ={arctan}\left({ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{arctan}\left(\:\frac{{tan}\left({u}_{{n}+\mathrm{1}} \right)\:−{tan}\left({u}_{{n}} \right)}{\mathrm{1}+{tan}\left({u}_{{n}} \right){tan}\left({u}_{{n}+\mathrm{1}} \right)}\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{N}} \:{arctan}\left({tan}\left({u}_{{n}+\mathrm{1}} −{u}_{{n}} \right)\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{N}} \:\left({u}_{{n}+\mathrm{1}} −{u}_{{n}} \right)=\:{u}_{{N}+\mathrm{1}} \:−{u}_{\mathrm{1}} ={arctan}\left({ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{N}+\mathrm{1}}\right)\right. \\ $$$$−{arctan}\left({ln}\left(\mathrm{2}\right)\right)\:{and}\: \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\:={arctan}\left({ln}\left(\mathrm{1}\right)\right)−{arctan}\left({ln}\left(\mathrm{2}\right)\right) \\ $$$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =−{arctan}\left({ln}\left(\mathrm{2}\right)\right)\:. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 26/Apr/18
the Q.is find Σ_(n=1) ^(+∞) arctan(((ln(1+(1/(n+1)))−ln(1+(1/n)))/(1+ln(1+(1/n))ln(1+(1/(n+1))))))
$${the}\:{Q}.{is}\:{find}\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:{arctan}\left(\frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}{\mathrm{1}+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)}\right) \\ $$

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