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Question Number 26812 by amit96 last updated on 30/Dec/17

sum of infinite seris tan^(−1) (2/n^2 )

$${sum}\:{of}\:{infinite}\:{seris} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}/{n}^{\mathrm{2}} \right) \\ $$

Answered by prakash jain last updated on 31/Dec/17

(2/n^2 )=(((n+1)−(n−1))/(1+(n+1)(n−1))) tan^(−1) (2/n^2 )=tan^(−1) (n+1)−tan^(−1) (n−1) 1: tan^(−1) 2−tan^(−1) 0 2: tan^(−1) 3−tan^(−1) 1 3: tan^(−1) 4−tan^(−1) 2 .. (n−1): tan^(−1) n−tan^(−1) (n−2) n: tan^(−1) (n+1)−tan^(−1) (n−1) sum of n terms tan^(−1) (n+1)+tan^(−1) (n)−tan^(−1) 1−tan^(−1) 0 Σ_(j=1) ^∞ tan^(−1) (2/n^2 )=(π/2)+(π/2)−(π/4)=((3π)/4)

$$\frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\frac{\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)}{\mathrm{1}+\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left({n}−\mathrm{1}\right) \\ $$$$\mathrm{1}:\:\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\mathrm{tan}^{−\mathrm{1}} \mathrm{0} \\ $$$$\mathrm{2}:\:\mathrm{tan}^{−\mathrm{1}} \mathrm{3}−\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$$\mathrm{3}:\:\mathrm{tan}^{−\mathrm{1}} \mathrm{4}−\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$$.. \\ $$$$\left({n}−\mathrm{1}\right):\:\mathrm{tan}^{−\mathrm{1}} {n}−\mathrm{tan}^{−\mathrm{1}} \left({n}−\mathrm{2}\right) \\ $$$${n}:\:\mathrm{tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \left({n}−\mathrm{1}\right) \\ $$$$\mathrm{sum}\:\mathrm{of}\:{n}\:\mathrm{terms} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left({n}\right)−\mathrm{tan}^{−\mathrm{1}} \mathrm{1}−\mathrm{tan}^{−\mathrm{1}} \mathrm{0} \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$

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