# Prove the equality: 1/1*2 + 1/2*3 + .. + 1/(n-1)*n = (n-1)/n

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### 2 Answers

We have to prove 1/1*2 + 1/2*3 + .. + 1/(n-1)*n = (n-1)/n. This can be done using induction. We first need to identify the first number for which the relation is true. Then for all following terms if the relation is true for n it should be true for n+1.

We see that for n = 1, the terms in the relation are not defined.

For n = 2, we have 1/(1*2*) = (2 - 1)/2 = 1/2

The expression is true for n = 2

Now let the expression be true for any n greater than 1.

For n + 1, we have

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

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

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

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

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

=> n^2/n*(n + 1)

=> [(n + 1) - 1]/[(n + 1)]

This shows that if the relation is true for any n>1, it is true for n+1.

**This proves that the relation holds for all n > 1.**

We'll apply the principle of mathematical induction to prove the equality.

We'll put P(n):1/1*2 + 1/2*3 + .. + 1/(n-1)*n = (n-1)/n

For n = 1, we'll have:

P(1): 1/1*2 = (1-1)/1

1/2 = 0

For n = 1, the principle does not hold.

We'll put n = 2:

P(2): 1/1*2 + 1/2*3 = (2-1)/2

P(2): 1/2 + 1/6 = 1/2

P(2): 2/3 = 1/2

For n = 2, the principle does not hold.

We'll put n = 3:

P(3): 1/1*2 + 1/2*3 = (3-1)/3

P(3): 2/3 = 2/3

Let P(n) to be true for n = k:

P(k): 1/1*2 + 1/2*3 + .. + 1/(k-1)*k = (k-1)/k

We'll verify if P(k+1) is true:

P(k+1):1/1*2 + 1/2*3 + .. + 1/(k-1)*k + 1/k(k+1) = k/(k+1)

But 1/1*2 + 1/2*3 + .. + 1/(k-1)*k = (k-1)/k = P(k), that is assumed to be true.

P(k+1): P(k) + 1/k(k+1) = k/(k+1)

P(k+1): (k-1)/k + 1/k(k+1) = k/(k+1)

P(k+1): (k^2 - 1 + 1)/k(k+1) = k/(k+1)

We'll eliminate like terms:

P(k+1): (k^2)/k(k+1) = k/(k+1)

We'll simplify and we'll get:

P(k+1): (k)/(k+1) = k/(k+1)

**Since P(k) is true, then P(k+1) is true, hence P(n) is true for any n>2.**