Given
limn--->
[r=1 ]
[ n] tan-1(1/2r2)
Firstly, to calculate the sum,
let S = [r=1 ][n ] tan-1(1/2r2)
= [r=1 ][n ] tan-1(2/4r2)
= [r=1 ][n ] tan-1(2/1+4r2-1)
= [r=1 ][n ] tan-1( (2r+1) -(2r-1)/ 1+ (2r+1)(2r-1) )
= [r=1 ][n ] tan-1 (2r+1) -tan-1 (2r-1)
On writing the values u get,
[ tan-13 -tan-11] +[ tan-1 5 -tan-1 3] + ...............+[ tan-1(2n+1) -tan-1 (2n-1)]
After all the terms get cancelled, u r left with
S= tan-1(2n+1) -tan-1 1
= tan-1 (2n/2n+2)
No taking the limit,
[n ]
[ ] tan-1 (n/n+1)
= [n ][ ] tan-1(1/1+ (1/n) )
On applying the limit, u get
= tan-1 1
= pi/4
Moderation Team
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