The series given is:

$2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \cdots + \frac{2}{3^{n-1}} + \cdots$

This is a geometric series where the first term $a = 2$ and the common ratio $r = \frac{1}{3}$.

Step 1: Formula for the $n$-th Partial Sum

The $n$-th partial sum $S_n$ of a geometric series with first term $a$ and common ratio $r$ (where $|r| < 1$) is given by:

$S_n = a \frac{1 - r^n}{1 - r}$

For this series:

• $a = 2$
• $r = \frac{1}{3}$

Substituting these values, the $n$-th partial sum $S_n$ becomes:

$S_n = 2 \frac{1 - \left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}}$

Simplifying further:

$S_n = 2 \frac{1 - \frac{1}{3^n}}{\frac{2}{3}} = 2 \cdot \frac{3}{2} \left(1 - \frac{1}{3^n}\right)$

$S_n = 3 \left(1 - \frac{1}{3^n}\right)$

Thus, the formula for the $n$-th partial sum $S_n$ is:

$S_n = 3 \left(1 - \frac{1}{3^n}\right)$

Step 2: Finding the Sum of the Series

If the series converges, the sum $S$ of an infinite geometric series is given by:

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

Since $|r| = \frac{1}{3} < 1$, the series converges. Therefore,

$S = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3$

Final Answer

1. The formula for the $n$-th partial sum $S_n$ of the series is:

   $S_n = 3 \left(1 - \frac{1}{3^n}\right)$

2. The sum of the series (if it converges) is:

   $S = 3$

Would you like additional details on the steps?

Here are some related questions:

1. How do you determine if a series converges?
2. What are common applications of geometric series in real-world problems?
3. How does the formula for a partial sum differ for arithmetic vs. geometric series?
4. What changes in the partial sum formula if the common ratio is negative?
5. How can you approximate the partial sum for a very large $n$?

Tip: When finding sums of geometric series, always check if the absolute value of the common ratio is less than 1 to confirm convergence.