The problem here is an infinite series:

$\sum_{n=1}^{\infty} \left( \frac{1}{1 + \sqrt{2}} \right)^{n-1}$

This is a geometric series with the first term $a = 1$ (since when $n = 1$, $\left( \frac{1}{1 + \sqrt{2}} \right)^0 = 1$) and the common ratio $r = \frac{1}{1 + \sqrt{2}}$.

The sum of an infinite geometric series $\sum_{n=0}^{\infty} ar^n$ is given by:

$\frac{a}{1 - r}$

Applying this formula:

1. Here, $a = 1$.
2. The common ratio $r = \frac{1}{1 + \sqrt{2}}$.

To simplify $1 - r$, we can rationalize the denominator of $r$ as follows:

$r = \frac{1}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{1 - 2} = \frac{\sqrt{2} - 1}{1}$

Thus, $r = \sqrt{2} - 1$, and $1 - r = 2 - \sqrt{2}$.

Now the sum becomes:

$\frac{1}{1 - (\sqrt{2} - 1)} = \frac{1}{2 - \sqrt{2}}$

To rationalize this, multiply the numerator and denominator by $2 + \sqrt{2}$:

$\frac{1 \cdot (2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{2 + \sqrt{2}}{4 - 2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{\sqrt{2}}{2}$

Thus, the final answer is:

$\boxed{\frac{9(1 + \sqrt{2})}{\sqrt{2}}}$

Would you like further clarification, or do you have additional questions?

Related Questions:

1. How do we recognize a geometric series in general?
2. Why is it important to rationalize the denominator when simplifying expressions?
3. Can this approach be applied to series with different terms and ratios?
4. How do you prove the convergence of a geometric series?
5. What are some applications of geometric series in real-world scenarios?

Tip: Always check if an infinite series is geometric before attempting to sum it; geometric series have straightforward formulas for convergence.