Question Number 210091 by mr W last updated on 30/Jul/24
$${find}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)=? \\ $$
Commented by AlagaIbile last updated on 30/Jul/24
![Σ_(n=1) ^∞ [tan^(−1) (2n+1)−tan^(−1) (2n−1)] Telescopic sum = tan^(−1) 1 = (π/4)](https://www.tinkutara.com/question/Q210093.png)
$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right)\right] \\ $$$$\:\:{Telescopic}\:{sum} \\ $$$$\:=\:\mathrm{tan}^{−\mathrm{1}} \:\mathrm{1} \\ $$$$\:=\:\frac{\pi}{\mathrm{4}} \\ $$
Commented by mr W last updated on 30/Jul/24
$${thanks}! \\ $$
Answered by mr W last updated on 30/Jul/24
![(1/(2n^2 ))=((2n+1−(2n−1))/(1+(2n+1)(2n−1))) say tan α=2n+1, tan β=2n−1 (1/(2n^2 ))=((tan α−tan β)/(1+tan α tan β))=tan (α−β) ⇒tan^(−1) ((1/(2n^2 )))=α−β=tan^(−1) (2n+1)−tan^(−1) (2n−1) Σ_(n=1) ^∞ tan^(−1) ((1/(2n^2 ))) =lim_(n→∞) Σ_(k=1) ^n tan^(−1) ((1/(2k^2 ))) =lim_(n→∞) [Σ_(k=1) ^n tan^(−1) (2k+1)−Σ_(k=1) ^n tan^(−1) (2k−1)] =lim_(n→∞) [{tan^(−1) 3+tan^(−1) 5+...+tan^(−1) (2n+1)}−{tan^(−1) 1+tan^(−1) 3+...+tan^(−1) (2n−1)}] =lim_(n→∞) [tan^(−1) (2n+1)−tan^(−1) 1] =(π/2)−(π/4) =(π/4) ✓](https://www.tinkutara.com/question/Q210096.png)
$$\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }=\frac{\mathrm{2}{n}+\mathrm{1}−\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{1}+\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${say}\:\mathrm{tan}\:\alpha=\mathrm{2}{n}+\mathrm{1},\:\mathrm{tan}\:\beta=\mathrm{2}{n}−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }=\frac{\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta}{\mathrm{1}+\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}=\mathrm{tan}\:\left(\alpha−\beta\right) \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right)=\alpha−\beta=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }\right) \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}{k}^{\mathrm{2}} }\right) \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{k}+\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{k}−\mathrm{1}\right)\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\left\{\mathrm{tan}^{−\mathrm{1}} \mathrm{3}+\mathrm{tan}^{−\mathrm{1}} \mathrm{5}+…+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)\right\}−\left\{\mathrm{tan}^{−\mathrm{1}} \mathrm{1}+\mathrm{tan}^{−\mathrm{1}} \mathrm{3}+…+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right)\right\}\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \mathrm{1}\right] \\ $$$$=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}} \\ $$$$=\frac{\pi}{\mathrm{4}}\:\checkmark \\ $$