In an exercise the following (simplified) step was made, now I can't recall how I came to this, please explain the following step:
$$... = \sum\limits_{m=0}^\infty \frac{1}{1000^m} = \frac{1}{1-\frac{1}{1000}} =... $$
In an exercise the following (simplified) step was made, now I can't recall how I came to this, please explain the following step:
$$... = \sum\limits_{m=0}^\infty \frac{1}{1000^m} = \frac{1}{1-\frac{1}{1000}} =... $$
Recall the known geometric series :
$$\frac{1}{1-w} = \sum_{n=0}^\infty w^{n} \quad \text{for} \quad |w|\leq1$$
Applying this to your given expression for $w = \frac{1}{1000}$ and for $n=m$ :
$$\sum\limits_{m=0}^\infty \frac{1}{1000^m} = \frac{1}{1-\frac{1}{1000}}$$
which holds, since :
$$\bigg|\frac{1}{1000}\bigg| < 1$$
This is an infinite geometric series and the sum to infinity of a geometric series is $\frac{a}{1-r}$, where a is the starting term which here is $\frac{1}{1000^0}=1$, and r is the ratio each term is multiplied by to get the next term which is $\frac{1}{1000}$.
Putting all this together we get $\frac{1}{1-\frac{1}{1000}}$
Hope that helps :)