Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 137721 by Ñï= last updated on 05/Apr/21

Σ_(n=1) ^∞ tan^(−1) (1/(2n^2 ))=? except use tan^(−1) (1/(2n^2 ))=tan^(−1) (1/(2n−1))−tan^(−1) (1/(2n+1)),any other way?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }=? \\ $$$${except}\:{use}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}},{any}\:{other}\:{way}? \\ $$

Answered by TANMAY PANACEA last updated on 05/Apr/21

tan^(−1) ((2/(4n^2 ))) =tan^(−1) ((2/(1+4n^2 −1))) =tan^(−1) ((((2n+1)−(2n−1))/(1+(2n+1)(2n−1)))) =tan^(−1) (2n+1)−tan^(−1) (2n−1) T_n =tan^(−1) (2n+1)−tan^(−1) (2n−1) T_1 =tan^(−1) (3)−tan^(−1) (1) T_2 =tan^(−1) (5)−tan^(−1) (3) T_3 =tan^(−1) (7)−tan^(−2) (5) ... ... T_n =tan^(−1) (2n+1)−tan^(−1) (2n−1) add them S_n =tan^(−1) (2n+1)−tan^(−1) (1) when n→∞ tan^(−1) (2n+1)→tan^(−1) (∞)=(π/2) so S_∞ =(π/2)−tan^(−1) (1) =(π/2)−(π/4)=(π/4)Tanmay

$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{4}{n}^{\mathrm{2}} }\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{1}+\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$${T}_{{n}} ={tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$${T}_{\mathrm{1}} ={tan}^{−\mathrm{1}} \left(\mathrm{3}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${T}_{\mathrm{2}} ={tan}^{−\mathrm{1}} \left(\mathrm{5}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{3}\right) \\ $$$${T}_{\mathrm{3}} ={tan}^{−\mathrm{1}} \left(\mathrm{7}\right)−{tan}^{−\mathrm{2}} \left(\mathrm{5}\right) \\ $$$$... \\ $$$$... \\ $$$${T}_{{n}} ={tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$${add}\:{them} \\ $$$${S}_{{n}} ={tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${when}\:{n}\rightarrow\infty \\ $$$${tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)\rightarrow{tan}^{−\mathrm{1}} \left(\infty\right)=\frac{\pi}{\mathrm{2}} \\ $$$${so}\:{S}_{\infty} =\frac{\pi}{\mathrm{2}}−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}}{Tanmay}\: \\ $$

Commented by Ñï= last updated on 06/Apr/21

thanks sir

$${thanks}\:{sir} \\ $$

Commented by TANMAY PANACEA last updated on 06/Apr/21

most welcome

$${most}\:{welcome} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com