SumFind the value of the sum 1/1*2+1/2*3+...+1/n(n+1).
We notice that the denominator of each fraction is the product of 2 consecutive numbers.
We'll decompose the fraction inot partial fractions:
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)