This expression represents a repeating decimal series with specific coefficients in each power of $10$. Let’s solve it step by step:

Given Series:

$1 + \frac{3}{10} + \frac{7}{10^2} + \frac{9}{10^3} + \frac{3}{10^4} + \frac{7}{10^5} + \frac{9}{10^6} + \ldots$

1. Pattern Identification:

   • The repeating coefficients are: $3, 7, 9$, and they cycle every 3 terms.
   • This series can be written in a compact way using powers of $10$: $S = 1 + \left(\frac{3}{10} + \frac{7}{10^2} + \frac{9}{10^3}\right) + \left(\frac{3}{10^4} + \frac{7}{10^5} + \frac{9}{10^6}\right) + \ldots$

2. Group and Simplify:

   • Factor out the repeating block $\frac{3}{10} + \frac{7}{10^2} + \frac{9}{10^3}$: $S = 1 + \left(\frac{3}{10} + \frac{7}{10^2} + \frac{9}{10^3}\right) \left(1 + \frac{1}{10^3} + \frac{1}{10^6} + \ldots\right)$

3. Simplify Infinite Geometric Series:

   • The repeating block sum is: $\frac{3}{10} + \frac{7}{10^2} + \frac{9}{10^3} = \frac{3}{10} + \frac{7}{100} + \frac{9}{1000} = \frac{300 + 70 + 9}{1000} = \frac{379}{1000}.$
   • The second term is an infinite geometric series: $1 + \frac{1}{10^3} + \frac{1}{10^6} + \ldots = \frac{1}{1 - \frac{1}{10^3}} = \frac{1}{1 - 0.001} = \frac{1}{0.999} = 1.001.$

4. Combine Everything:

   • Substitute back: $S = 1 + \frac{379}{1000} \times 1.001.$
   • Multiply: $\frac{379}{1000} \times 1.001 = \frac{379.379}{1000}.$
   • Add to $1$: $S = 1 + 0.379379 = 1.379379.$

Final Answer:

$S = 1.379379\ldots = 1.\overline{379}.$

Would you like more details or a verification of this result?

Related Questions:

1. How do geometric series help in solving infinite decimal expansions?
2. Can you explain why the repeating pattern contributes to a rational number?
3. What other forms can $1.\overline{379}$ be expressed as?
4. Can this result be verified by converting $1.\overline{379}$ back into a fraction?
5. How would the result change if the repeating block were longer?

Tip:

Always look for repeating patterns and group terms systematically to simplify infinite series problems!