# why $\sum_{k=0}^{\infty}(10^{-2})^k = \frac{1}{1-10^{-2}}$

i was reading a book and suddenly saw this step: $\sum_{k=0}^{\infty}(10^{-2})^k = \frac{1}{1-10^{-2}}$
i am actually not bad at calculation and also i am okay in precalculus, but i am really stuck here, not knowing how come this equality is true. i cannot see steps done between them. please dont downvote if the question is too simple. i cannot even give you any of my steps, cos i didnot do any.

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It’s a geometric series; see if the linked article helps. –  Brian M. Scott Dec 25 '12 at 21:56
oh i got it, isnot cos of geometric series? oh yeah, Brian, simultanuous thoughts.. :) –  doniyor Dec 25 '12 at 21:56
Yes, you got it. :-) –  Brian M. Scott Dec 25 '12 at 21:56

Recall the geometric series $$\sum_{k=0}^{\infty} x^k = \dfrac1{1-x} \,\,\,\,\,\,\, \forall x \in (-1,1)$$ Since $10^{-2} \in (-1,1)$, we have that $$\sum_{k=0}^{\infty} \left(10^{-2} \right)^k = \dfrac1{1-10^{-2}}$$

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yes, absolutely, thanks man, sorry for the dumbness! :D –  doniyor Dec 25 '12 at 21:58
Moreover, the same formula holds for all $x \in \mathbb{C}$ with $|x|<1$. –  Fly by Night Dec 25 '12 at 23:00

What you have here is a geometric series. The first term, denoted by $a$, is $(10^{-2})^0=1$ while the common ratio, denoted by $r$, is $10^{-2} = 1/100$. We are interested in finding the sum of all of the terms. Let us use $S$ to denote $a + ar + ar^2 + \cdots$ There is a well-known formula for $S$. Let's define:

$$\begin{array}{ccc} S & = & 1 + 10^{-2} + 10^{-4} + 10^{-6} + 10^{-8} + \cdots \, \end{array}$$

A nice little trick to employ if to compare $S$ and $r \times S$. We have: $$\begin{array}{ccc} S & = & 1 + 10^{-2} + 10^{-4} + 10^{-6} + 10^{-8} + \cdots \\ 10^{-2} \times S & = & 10^{-2} + 10^{-4} + 10^{-6} + 10^{-8} + 10^{-10} + \cdots \end{array}$$

The cunning step is to consider $S - 10^{-2}S = (1-10^{-2})S$. This gives:

$$(1-10^{-2})S = 1 \iff S = \frac{1}{1-10^{-2}} \, .$$

Moreover:

$$S = \frac{1}{1-10^{-2}} = \frac{1}{\frac{100}{100}-\frac{1}{100}} = \frac{1}{\frac{99}{100}} = \frac{100}{99} = 1.\overline{01} \, .$$

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+1 Nice...${}{}{}{}{}$ –  amWhy Dec 25 '12 at 22:46
@amWhy Thanks for the kind words. I can't help thinking that it looks ugly using $10^{-2}$ in place of $r$, but hey. All the best. –  Fly by Night Dec 25 '12 at 22:59
I think it's wonderful for having explained the logic behind the $\frac{1}{1 - r}$ formula, rather than just citing it as a "rule" to follow! And for showing how that relates to this problem! –  amWhy Dec 25 '12 at 23:02
Thanks @Fly, great explanation! –  doniyor Dec 25 '12 at 23:52