The question asks to approximate the infinite sum

$\sum_{n=1}^{\infty} \frac{(1)^n}{10^n}$

correct to 4 decimal places.

This series can be identified as a geometric series where the general term is

$a_n = \frac{(1)^n}{10^n} = \frac{1}{10^n},$

with the first term $a = \frac{1}{10}$ and common ratio $r = \frac{1}{10}$.

For a geometric series with $|r| < 1$, the sum of the infinite series is given by:

$S = \frac{a}{1 - r}.$

Substituting the values:

$a = \frac{1}{10}, \quad r = \frac{1}{10}.$

$S = \frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9}.$

Calculating $S$:

$S = 0.1111 \, (\text{correct to 4 decimal places}).$

The approximation of the series is $\mathbf{0.1111}$.

Let me know if you have any questions or need further clarification!

Here are 5 related questions to expand your understanding:

1. What would the sum be if the common ratio $r$ changed to $\frac{1}{2}$?
2. How would you approximate this series using partial sums?
3. What happens to the series' sum if $r \geq 1$?
4. Can you apply this formula to non-geometric series?
5. How does rounding affect the accuracy of the solution?

Tip: Always check if the series converges (e.g., $|r| < 1$) before applying the infinite sum formula!