Mathematics in Ancient and Medieval India c8.s3.133

c8.s3.133

Now making use of the result, $_n → ∞ 1/(np+1) _r=0n-1 rp = 1/(p+1)$ We obtain, $S = t - (t³)/3 + + (-1)ν-1 (t2ν-1)/(2ν-1) + (-1)ν R_ν$ where 0 < R_ν < 1/(2ν+1)