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sum of the series upto infinity from 0 to infinitytan-1(1/1+1+12)+tan-1(1/1+2+22)+tan-1(1/1+3+32).................... |
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Answer» The general term of the expression is tr. = tan^-1(1/(1+r+r^^2)) =>tr =tan^-1(r+1)-tan^-1(r) So inserting r=1 to infinity and summing up we get the required sum as S =t1+t2+t3+.......tinfinity =tan^-1(infinity)-tan^-1(1) =pi/2-pi/4=pi/4 |
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