\( 1 + \frac { 10 } { 11 } + \frac { 100 } { 121 } \ldots \)

The series given is \[ S = 1 + \frac{10}{11} + \frac{100}{121} + \ldots \] We can observe that the terms can be expressed in a more general form. The \( n \)-th term of the series can be written as: \[ \frac{10^{n-1}}{11^{n-1}} = \left( \frac{10}{11} \right)^{n-1} \] Thus, we can rewrite the series \( S \) as: \[ S = \sum_{n=1}^{\infty} \left( \frac{10}{11} \right)^{n-1} \] This is a geometric series with the first term \( a = 1 \) (when \( n=1 \)) and the common ratio \( r = \frac{10}{11} \). The sum of an infinite geometric series can be calculated using the formula: \[ S = \frac{a}{1 - r} \] provided that \( |r| < 1 \). In our case, \( a = 1 \) and \( r = \frac{10}{11} \), which satisfies \( |r| < 1 \). Now, substituting the values into the formula: \[ S = \frac{1}{1 - \frac{10}{11}} = \frac{1}{\frac{1}{11}} = 11 \] Thus, the sum of the series is \[ \boxed{11} \]