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find the sum of the series tan ​-1 (1/3) +tan ​-1 (1/7)+tan ​-1 (1/13)+.....+tan ​-1 (1/(n ​2 +n+1)+...+..infinity

find the sum of the series tan​-1(1/3) +tan​-1(1/7)+tan​-1(1/13)+.....+tan​-1(1/(n​2+n+1)+...+..infinity
 
 

Grade:11

2 Answers

Akshit Patel
26 Points
6 years ago
The answer is π/4. Its general term is tan-1(1/n2+n+1).And we can write it as tan-1((n+1)-n)/1+n(n+1)) =tan-1(n+1) - tan-1 (n)When we sum it from n=1 to infinity we get tan-1(infinity)-tan-1 (1)=(π/2)-(π/4)=π/4
Muztaba
13 Points
6 years ago
This is basically summation(S) of 1/(r^2+r+1) over r, from 1 to infinity. Let Tr be the rth term, Tr=tan(-1) [1/(r^2+r+1)] =tan(-1) [((r+1)-r)/(1+ (r+1)r)] =tan(-1) (r+1) - tan(-1)(r)Now we find T1, T2,.., Tn,... and add them together. T1= tan(-1)2 - tan(-1)1T2=tan(-1)3 - tan(-1)2........Tn=tan(-1)(n+1) - tan(-1)(n)...... _____________________________S= - tan(-1)1 [since cross cancellation takes place in the summation] As n tends to infinity, we get S= - pi/4**tan(-1)x mean tan inverse of x.

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