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 05-05-2006 #4 Shura     Join Date: Mar 2006 Posts: 12 Another Solution I asked a teacher about this today and she pointed me in the direction of using a geometric progression to solve / prove this. (Although I not sure how rigorous this is poof wise) S(n) represents S with an n subscript, but I cannot get subscripts to work. Let S(n) = Sum of the first n terms of a geometric progression. a is the first term and r is the common ratio. S(n) = a + ar + ar2 + ar3 + ... + arn-1 Multiply everything by r r . S(n) = ar + ar2 + ar3 + ... + arn-1 + arn Subtract this from the original expression, apart from the first term in one expression and the last in the other all of the terms cancel. r . S(n) - S(n) = arn - a Simplify S(n)(r-1) = a (rn - 1) S(n) = a(rn - 1) / (r - 1) S(n) = a(1 - r n) / (1 - r) The reason I have just done all of this (I did have a reason ) , is because 0.9 recurring can be written as a geometric progression of nine tenths, plus 9 hundreths, plus nine thousandths etc. So S(n) = 9/10 + 9/100 + 9/1000 + ... + 9/10n common ratio r = 1/10 You can calculate this by comparing the ratio of two consecutive terms. e.g. 0.09 / 0.9 = 0.1 For |r| < 1 as n -> infinity, rn -> 0 This should read for values of r between -1 and 1 (Otherwise the series doesn't converge to a value), as n tends towards (approaches) infinity rn approaches zero. (You may need to look up limits, this is the tricky bit ) So a(1 - r n) / (1 / r) becomes a/(1-r) for the sum to infinity As a = 9/10 and r = 1/10 S(infinity) = a/(1-r) = (9/10) / (1 - 1/10) S(infinity) would be written S with the infinity symbol as a subscript. = (9/10) / (9/10) = 1 So point 9 recurring = 1. Isn't Mr Hui's way simplier. - Shura __________________ "Mathematics is the gate and key to the sciences." -- Roger Bacon