Compute the sum \(\displaystyle \frac{1}{2^2 - 1} + \frac{1}{3^2 - 1} + \frac{1}{4^2 - 1} + \dots + \frac{1}{20^2 - 1}\)
+128448 1 1 1 1
____________ + ___________ + ____________ + .........+ ______________ =
(2 + 1) (2 - 1) (3 + 1) (3 -1) (4 + 1) ( 4 -1) ( 20 + 1) (20 - 1)
By partial fractions we have
1 A B
____________ = _____ + _____
( n + 1) (n - 1) n + 1 n - 1
1 = A(n - 1) + B( n + 1)
1 = An - A + Bn + B
1 = (A + B)n + (B - A)
A + B =0
-A + B = 1
2B = 1
B = 1/2
A = -1/2
So we have that
1 1 1 1
____________ = _______ - ________ = ____ [ 1/(n -1) - 1/( n+1) ]
(n + 1) ( n - 1) 2(n -1) 2 ( n + 1) 2
So we have
(1/2) [ (1/1 - 1/3) + (1/2 - 1/4) + ( 1/3 - 1/5) + ( 1/4 - 1/6) + ( 1/5 - 1/7) + .......+ (1/18 - 1/20) + (1/19 - 1/21) ]
The terms in red will "cancel"
We are left with
(1/2) ( 1 + 1/2 - 1/20 - 1/21) =
(1/2) ( 3/2 - (1/20 + 1/21) ) =
(1/2)(3/2 - ( 41/420) ] =
(1/2) ( 630 - 41) / 420 =
589 / 840
