What is the value of \( 0.1+0.11+0.111+\cdots+0.1111111111 ? \)

To solve the sum \( 0.1 + 0.11 + 0.111 + \cdots + 0.1111111111 \), we first express each term in a more manageable form. The first term is: \[ 0.1 = \frac{1}{10} \] The second term is: \[ 0.11 = \frac{11}{100} = \frac{1}{10} + \frac{1}{100} \] The third term is: \[ 0.111 = \frac{111}{1000} = \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} \] Continuing this pattern, we observe that each term \( 0.111\ldots1 \) (with \( n \) ones) can be written as: \[ 0.111\ldots1 = \sum_{k=1}^{n} \frac{1}{10^k} \] The sum \( 0.111\ldots1 \) then can be rewritten as follows for \( n \) ones: \[ 0.111\ldots1 = \frac{1}{10} + \frac{1}{100} + \cdots + \frac{1}{10^n} \] This is a geometric series with the first term \( a = \frac{1}{10} \) and the common ratio \( r = \frac{1}{10} \). The sum of the first \( n \) terms of a geometric series can be calculated using the formula: \[ S_n = a \frac{1 - r^n}{1 - r} \] Substituting the values, we get: \[ S_n = \frac{1/10 (1 - (1/10)^n)}{1 - 1/10} = \frac{1/10 (1 - 1/10^n)}{9/10} = \frac{1 - 1/10^n}{9} \] Now, for the original problem, we must sum this for \( n = 1 \) to \( 10 \): \[ S_{10} = 0.1 + 0.11 + 0.111 + \cdots + 0.1111111111 = \sum_{n=1}^{10} \left( \frac{1 - \frac{1}{10^n}}{9} \right) \] This can be separated into two distinct sums: \[ S_{10} = \frac{1}{9} \sum_{n=1}^{10} (1 - \frac{1}{10^n}) = \frac{1}{9} \left( \sum_{n=1}^{10} 1 - \sum_{n=1}^{10} \frac{1}{10^n} \right) \] The first sum, \( \sum_{n=1}^{10} 1 \), gives us \( 10 \) since there are 10 terms: \[ \sum_{n=1}^{10} 1 = 10 \] The second sum \( \sum_{n=1}^{10} \frac{1}{10^n} \) is another geometric series: \[ \sum_{n=1}^{10} \frac{1}{10^n} = \frac{1/10 (1 - (1/10)^{10})}{1 - 1/10} = \frac{1/10 (1 - \frac{1}{10^{10}})}{9/10} = \frac{1 - \frac{1}{10^{10}}}{9} \] Putting it all together, we have: \[ S_{10} = \frac{1}{9} \left( 10 - \frac{1 - \frac{1}{10^{10}}}{9} \right) = \frac{10}{9} - \frac{1 - \frac{1}{10^{10}}}{81} \] \[ = \frac{10}{9} - \frac{81 - 1/10^{10}}{81} \] Now simplifying: \[ = \frac{10 \cdot 81 - (81 - \frac{1}{10^{10}})}{729} = \frac{810 - 81 + \frac{1}{10^{10}} }{729} = \frac{729 + \frac{1}{10^{10}}}{729} \] Taking the limit as \( n \) approaches 10, the fraction effectively simplifies as \( \frac{729}{729} = 1 \). Thus, the final