How does the minus come in Geometric Series

Question

I'm looking into the geometric series and can't understand how the 1 - .01 comes in below:

0.272727... = 0.27 + 0.0027 + 0.000027 + 0.00000027 + ...
             = 0.27 + 0.27(.01) + 0.27(.01)^2 + 0.27(.01)^3 + ...
             = 0.27 / (1-.01)
             = 0.27 / 0.99
             = 27/99
             = 3/11

Answer 1

This comes from a general formula for how you evaluate a geometric series:

$$\sum_{n=0}^{\infty} ax^n = \frac{a}{1-x},$$

which holds when $|x|<1$.

For a proof of this, you might want to check out this wiki-article: http://en.wikipedia.org/wiki/Geometric_progression#Derivation

This shows you how the "$-1$" pops up for a geometric terms with a finite number of terms, and it will be similar (by going to the limit) for an infinite number of terms. Hope this helps.

Answer 2

By induction it's easy to prove that

$$\forall\,a,b\in\Bbb R\;,\;\;\forall\,n\in\Bbb N\;:\;\;(a^n-b^n)=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})$$

From here we get the formula for the sum of a finite geometric progression:

$$\sum_{k=0}^naq^k=a(1+q+q^2+\ldots+q^n)=a\frac{1-q^{n+1}}{1-q}$$

Now, we have that $\;|q|<1\implies q^n\xrightarrow[n\to\infty]{}0\;$ , so now put all together and get your answer.

Answer 3

Question is why do solve it using geometric progressions at all. When you use the basic method taught as kids in school, you're basically deriving the sum of an infinite GP unknowingly. Here's the easiest way-

x=0.272727272727... And 100x=27.272727272727.... Thus, 99x=27 x=27/99

This is basically like deriving the sum of an infinite gp, only instead of a and r, we're using numbers here!