AL Pure Maths (Partial)

(a) Let [x(x+1)]^-1 = a/x + b /(x+1)
a(x+1) + bx = 1
Put x = 0,
a=1
then b = -1.
therefore, [x(x+1)]^-1 = 1/x - 1/(x+1)

(b) Denote nSx be summation from r=1 to r=n, where r is a dummy variable.
Take summation from r = 1 to n in (a),
then (n-1)S2 [1/(r^2)] <= nS1 [1/r(r+1)]
= nS1 [1/x - 1/(x+1)] from (a)
= 1- 1/(n+1)
(n-1)S1 [1/(r^2)] <= 1 - 1/(n+1) + 1
1/(n+1) + (n-1)S1 [1/(r^2)] <= 2
1/(n+1)^2 + (n-1)S1 [1/(r^2)] <= 2
nS1 [1/(r^2)] <= 2
which is the result required.
As the sequence nS1 [1/(r^2)] is divergent,
so the limit does not exist.

I hope the expression in the answer is clear enough.

2009-11-11 20:19:06 補充：
< = <--

2009-11-11 20:19:41 補充：
"<" is "<"

2009-11-11 20:20:19 補充：
&1t; is