Shoutbox

Hello forum people who know more math than me (probably everyone ;o)

I'm trying to solve this question, but I'm kind of lost and have not a clue what to do to prove it



so could you kind people please help

1 / n(n+1) = 1/n - 1/(n+1) for any n >=1.

Replacing all terms we get:
1/1 - 1/2 + 1/2 - 1/3 + ... + 1/n - 1/(n+1)

All terms cancel out in pairs, except first and last:
1/1 - 1/(n+1)

Which is what we need.

Induction is so much better

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Originally posted by markee in chat with -dt-, plus some more

1/(1*2) = 1 - 1/(1+1); n=1
1/2 = 1 - 1/2
1/2 = 1/2

true for n=1

assume true for n=k

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

show true for P(k+1):

LHS = 1/(1*2) + 1/(2*3) + ... + 1/(k-1)k + 1/k(k+1)
= 1 - 1/(k+1) + 1/(k+1)(k+2)
= 1 - [ (k+1)(k+2) - (k+1) ]/(k+1)(k+1)(k+2)    ---> it is -(k+1) because of the negative at the front of the fraction
= 1 - [k+2-1] / (k+1)(k+2)
= 1 - (k+1) / (k+1)(k+2)
= 1 - 1/(k+2)
= RHS

let k=1 as it is true for this value.  Therefore true for all integers above 1.

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Yay mathematical induction. A more formal conclusion:

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Since it is true for n = 1 it must also be true for n = 1+1 = 2, and since it is true for n = 2 it must also be true for n = 2+1 = 3, and so on for all integers n >= 1

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