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

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### 1 Answer

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.**