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Question Number 27189 by abdo imad last updated on 02/Jan/18

let give S_(n ) = Σ_(p=1) ^(p=n)  arctan ((1/(2p^2 )) )  find lim_(n−>∝)  S_n   .

$${let}\:{give}\:{S}_{{n}\:} =\:\sum_{{p}=\mathrm{1}} ^{{p}={n}} \:{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\:\right)\:\:{find}\:{lim}_{{n}−>\propto} \:{S}_{{n}} \:\:. \\ $$

Answered by prakash jain last updated on 03/Jan/18

tan^(−1) (1/(2p^2 ))=tan^(−1) (((2p+1)−(2p−1))/(1+(2p+1)(2p+1)))  =tan^(−1) (2p+1)−tan^(−1) (2p−1)  p:1 → tan^(−1) 3−tan^(−1) 1  p:2→tan^(−1) 5−tan^(−1) 3  ⋮  p:n−1→tan^(−1) (2n−1)−tan^(−1) (2n−3)  p:n→tan^(−1) (2n+1)−tan^(−1) (2n−3>1)  Σ_(p=1) ^n tan^(−1) (1/(2p^2 ))=tan^(−1) (2n+1)−tan^(−1) 1  lim_(n→∞) Σ_(p=1) ^n tan^(−1) (1/(2p^2 ))=(π/2)−(π/4)=(π/4)

$$\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} \frac{\left(\mathrm{2}{p}+\mathrm{1}\right)−\left(\mathrm{2}{p}−\mathrm{1}\right)}{\mathrm{1}+\left(\mathrm{2}{p}+\mathrm{1}\right)\left(\mathrm{2}{p}+\mathrm{1}\right)} \\ $$ $$=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}−\mathrm{1}\right) \\ $$ $${p}:\mathrm{1}\:\rightarrow\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3}−\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$ $${p}:\mathrm{2}\rightarrow\mathrm{tan}^{−\mathrm{1}} \mathrm{5}−\mathrm{tan}^{−\mathrm{1}} \mathrm{3} \\ $$ $$\vdots \\ $$ $${p}:{n}−\mathrm{1}\rightarrow\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{3}\right) \\ $$ $${p}:{n}\rightarrow\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{3}>\mathrm{1}\right) \\ $$ $$\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$ $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\ $$

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