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 + ar**^{2} + ar^{3} + ... + ar^{n-1}
Multiply everything by

**r**
**r . S(n) = ar + ar**^{2} + ar^{3} + ... + ar^{n-1} + ar^{n}
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) = ar**^{n} - a
Simplify

**S(n)(r-1) = a (r**^{n} - 1)

S(n) = a(r^{n} - 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 + ... + 9/10**^{n}
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**,

**r**^{n} -> 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 r

^{n} 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