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.  
 A: 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:
$$S = \frac{1}{1-10^{-2}} = \frac{1}{\frac{100}{100}-\frac{1}{100}} = \frac{1}{\frac{99}{100}} = \frac{100}{99} = 1.\overline{01} \, .$$
A: 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
$$\sum_{k=0}^{\infty} \left(10^{-2} \right)^k = \dfrac1{1-10^{-2}}$$
