IF I(r)=r(r²-1),then find ∑^∞_r=2 1/I(r).

1/(r(r^2-1) = 1/(r-1)(r)(r+1) Numerator is 1 = ( (r+1) - (r-1) )/2 Now the series is: 1/2( 1/(r-1)(r) - 1/(r)(r+1) ) 1st term: 1/2 ( 1/1.2 - 1/2.3 ) 2nd term: 1/2 ( 1/2.3 - 1/3.4) 3rd term: 1/2 ( 1/3.4 - 1/4.5) . . . . nth term: 1/2 ( 1/(n-1)n - 1/n(n+1)) On adding term cancel out. The summation gives 1/2( 1/1.2 - 1/n(n+1)) as n goes infinity: Ans: 1/4 Thanks and Regards, Pratik Tibrewal askiitians faculty BTech IITG