InterviewSolution
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InterviewSolution
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`Sigma_(r=1)^(infty) tan^(-1)(1)/(2r^(2))=t` then tan t is equal toA. `2/3`B. 1C. `sqrt(5)/(3)`D. none of these |
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Answer» we have `t=underset(oo)overset(r=1)Sigma tan^(-1)((1)/(2r)^(2))` `rarr t=underset(r=1)overset(oo)Sigma tan^(-1)(2)/(1+4r^(2)-1)` `underset(r=1)overset(oo)Sigma{tan^(-1)(2r+1)-tan^(-1)(2r-1)}` `=underset(n rarr oo) lim {(tan^(-1) 3-tan^(-1)(2n+1)-tan^(-1)(2r-1)}` `=underset(n rarr oo) lim {(tan^(-1) 3-tan^(-1)+(tan^(-1)5-tan^(-1)3)+(tan^(-1)(2n+1)-tan^(-1)(2n-1)}` `rarr t= underset(n rarr oo) lim (tan^(-1)(2n+1)-tan^(-1)1)=(pi)/(2)-(pi)/(4)=(pi)/(4)` `therefore tan t=1` |
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