What is the value of the sum 1/1*2+1/2*3+...+1/n(n+1) ?

We notice that the denominator of each ratio is the
product of 2 consecutive numbers.

We can write this kind of
ratio 1/n(n+1) as the result of addition or subtraction of 2 elementary
fractions:

We'll write the
ratio:

1/n(n+1) = A/n + B/(n+1)
(1)

We'll multiply the ratio A/n by (n+1) and we'll
multiply the ratio B/(n+1) by n.

1/n(n+1) = [A(n+1) +
Bn]/n(n+1)

Since the denominators of both sides are
matching, we'll write the numerators, only.

1 = A(n+1) +
Bn

We'll remove the
brackets:

1 = An + A +
Bn

We'll factorize by n to the right
side:

1 = n(A+B) + A

If the
expressions from both sides are equivalent, the correspondent coefficients are
equal.

A+B = 0

A =
1

1 + B = 0

B =
-1

We'll substitute A and B into the expression
(1):

1/n(n+1) = 1/n - 1/(n+1)
(2)

According to  (2), we'll give values to
n:

for n = 1 => 1/1*2 = 1/1 -
1/2

for n = 2 => 1/2*3 = 1/2 -
1/3

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

for
n = n-1 => 1/n(n-1) = 1/(n-1) - 1/n

for n = n
=> 1/n(n+1) = 1/n - 1/(n+1)

If we'll add the ratios
from the left side, we'll get:

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

We'll eliminate like terms from the right
side:

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

S = (n + 1 -
1)/(n+1)

We'll eliminate like terms and we'll get the final
result of the sum:

S = n
/(n+1)