Question Number 181238 by SANOGO last updated on 23/Nov/22
$${calculer} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+{oo}} {\sum}}{artan}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right) \\ $$
Answered by qaz last updated on 23/Nov/22
![arctan (2/n^2 )=arctan (n+1)−arctan (n−1) Σarctan (2/n^2 )=Σa_n =Σa_(2n−1) +Σa_(2n) =Σ{[arctan (2n)−arctan (2n−2)]+[arctan (2n+1)−arctan (2n−1)]} =[arctan (2∙+∞)−arctan (2∙1−2)]+[arctan (2∙+∞+1)−arctan (2∙1−1)] =(3/4)π](Q181241.png)
$$\mathrm{arctan}\:\frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\mathrm{arctan}\:\left({n}+\mathrm{1}\right)−\mathrm{arctan}\:\left({n}−\mathrm{1}\right) \\ $$$$\Sigma\mathrm{arctan}\:\frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\Sigma{a}_{{n}} =\Sigma{a}_{\mathrm{2}{n}−\mathrm{1}} +\Sigma{a}_{\mathrm{2}{n}} \\ $$$$=\Sigma\left\{\left[\mathrm{arctan}\:\left(\mathrm{2}{n}\right)−\mathrm{arctan}\:\left(\mathrm{2}{n}−\mathrm{2}\right)\right]+\left[\mathrm{arctan}\:\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{arctan}\:\left(\mathrm{2}{n}−\mathrm{1}\right)\right]\right\} \\ $$$$=\left[\mathrm{arctan}\:\left(\mathrm{2}\centerdot+\infty\right)−\mathrm{arctan}\:\left(\mathrm{2}\centerdot\mathrm{1}−\mathrm{2}\right)\right]+\left[\mathrm{arctan}\:\left(\mathrm{2}\centerdot+\infty+\mathrm{1}\right)−\mathrm{arctan}\:\left(\mathrm{2}\centerdot\mathrm{1}−\mathrm{1}\right)\right] \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\pi \\ $$
Commented by mr W last updated on 23/Nov/22
$${seems}\:{to}\:{be}\:{wrong}.\:{please}\:{recheck}\:{sir}. \\ $$
Commented by qaz last updated on 23/Nov/22
$$\mathrm{0}<\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\:\frac{\mathrm{2}}{{n}^{\mathrm{2}} }<\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\approx\mathrm{3}.\mathrm{29} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\pi\approx\mathrm{2}.\mathrm{35}\:\:\:\:\:,\:\frac{\mathrm{5}}{\mathrm{4}}\pi\approx\mathrm{3}.\mathrm{92} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\pi\:{is}\:{correct},{i}\:{think} \\ $$
Commented by MJS_new last updated on 23/Nov/22
$$\mathrm{I}\:\mathrm{also}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct} \\ $$