[Shura]

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 + ... + ar^n-1

Multiply everything by r

r . S(n) = ar + ar² + ar³ + ... + ar^n-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