Note that we can write
1 1
__________ = _______________
n ( n^2 - 1) n ( n + 1) (n - 1)
Using partial fractions
1 A B C
___________ = ___ + _____ + _____ multiply through by n(n + 1)(n -1)
n(n + 1) (n-1) n n + 1 n - 1
1 = A( n + 1) ( n - 1) + Bn(n - 1) + Cn ( n + 1)
1 = A( n^2 - 1) + B( n^2 - n) + C(n^2 + n)
1 = (A + B + C)n^2 + (B - C)n - A
Equating coefficients
A = -1
B - C = 0
A + B + C = 0
Add the second and third equations
A + 2B = 0
B = 1/2
C = 1/2
So we have
( -1/n + 1/[2(n + 1)] + 1'/[ 2 ( n - 1) ] =
Sum n = 2 to n = 100
(-1/2 + 1/ 6 + 1/2) +
(-1/3 + 1/8 + 1/4) +
(-1/4 + 1/10 + 1/6) +
(-1/5 + 1/12 + 1/8) +
(-1/6 + 1/14 + 1/10) +
(-1/7 + 1/16 + 1/12)
(-1/8 + 1/18 + 1/14) + ........+
(-1/97 + 1/196 + 1/192) +
(-1/98 + 1/198 + 1/194) +
(-1/99 + 1/200 + 1/196) +
( -1/100 + 1/202 + 1/198) =
All the terms in red will " cancel" and we are left with
(1/4 + 1/200 + 1/202 - 1/100)
(50/ 200 + 1/200 + 1/202 - 2/200) =
(49/200 + 1/202) =
[ 49 (202 ) + 200 ] 10098 5049
_______________ = _________ = _______
200 * 202 40400 20200