To evaluate the infinite sum
\[ \sum_{i=1}^{\infty} 100 \left( \frac{9}{10} \right)^{i-1}, \]
we can recognize that this is an infinite geometric series. In a geometric series, the general form is
\[ \sum_{i=0}^{\infty} ar^i, \]
where \(a\) is the first term and \(r\) is the common ratio. The formula for the sum of an infinite geometric series, when \(|r| < 1\), is given by:
\[ S = \frac{a}{1 - r}. \]
In this specific case:
- The first term (when \(i=1\)) is \(100\) (which corresponds to \(i-1=0\)).
- The common ratio \(r\) is \(\frac{9}{10}\).
Now we can apply the infinite geometric series sum formula:
- The first term \(a\) is \(100\).
- The common ratio \(r\) is \(\frac{9}{10}\).
We can plug these into the formula:
\[ S = \frac{100}{1 - \frac{9}{10}} = \frac{100}{\frac{1}{10}} = 100 \times 10 = 1000. \]
Thus, the value of the infinite sum is
\[ \boxed{1000}. \]