Evaluate the sum 1/1*3+1/3*5+...+1/(2n-1)(2n+1).

We notice that we can write the k-th term of the sum
as:

1/(2k-1)(2k+1)

We'll decompose
the term in a sum of elementary fractions:

1/(2k-1)(2k+1) =
A/(2k-1) + B/(2k+1)

We'll factorize by (2k-1)(2k+1) both
sides:

1 = A(2k+1) +
B(2k-1)

We'll remove the
brackets:

1 = 2Ak + A + 2Bk -
B

1 = 2k(A+B) +
(A-B)

Comparing both sides, we'll
get:

A+B = 0 => A =
-B

A-B = 1

A = 1 +
B

1 + 2B = 0

B = -1/2
=> A = 1/2

1/(2k-1)(2k+1) = 1/2(2k-1) -
1/2(2k+1)

Now, we'll calculate the
sum:

k = 1=> 1/1*3 = 1/2*1 -
1/2*3

k = 2 => 1/3*5 = 1/2*3 -
1/2*5

......................................

k
= n => 1/(2n-1)(2n+1) = 1/2*(2n-1) -
1/2*(2n+1)

We'll add the terms, we'll eliminate the like
ones and we'll get:

1/1*3 +...+ 1/(2n-1)(2n+1) = (1/2)[1 -
1/(2n+1)]

1/1*3 +...+ 1/(2n-1)(2n+1) = (1/2)[(2n + 1 -
1)/(2n+1)]

We'll simplify and we'll
get:

1/1*3 +...+ 1/(2n-1)(2n+1) =
n/(2n+1)