The given series appears to be:

$S = -1 + \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \dots + (-1)^n \frac{1}{2^{n-1}} + \dots$

This is an alternating series with a general term:

$a_n = (-1)^n \frac{1}{2^{n-1}}$

Step 1: Calculate the partial sum

The $n$-th partial sum is the sum of the first $n$ terms:

$S_n = -1 + \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \dots + (-1)^n \frac{1}{2^{n-1}}$

We can write this series as a geometric series by factoring out the powers of $\frac{1}{2}$.

Step 2: Express as a geometric series

Factor out $-1$ and write:

$S = -1 + \sum_{n=1}^\infty (-1)^n \frac{1}{2^n}.$

Here, the geometric series inside is:

$\sum_{n=1}^\infty (-1)^n \frac{1}{2^n}.$

For a geometric series of the form:

$\sum_{n=0}^\infty ar^n = \frac{a}{1-r}, \quad |r| < 1,$

we compute $r = -\frac{1}{2}$ and $a = -\frac{1}{2}$. The partial sum becomes:

$\sum_{n=1}^\infty (-1)^n \frac{1}{2^n} = \frac{-\frac{1}{2}}{1 - \left(-\frac{1}{2}\right)} = \frac{-\frac{1}{2}}{1 + \frac{1}{2}} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{3}.$

Step 3: Combine the results

Adding back the initial term $-1$, the sum of the series is:

$S = -1 + \left(-\frac{1}{3}\right) = -\frac{4}{3}.$

Conclusion

The series converges to:

$S = -\frac{4}{3}.$

Would you like a detailed explanation of any part, or have any other questions? Here are some related topics to explore:

1. How does an alternating series test ensure convergence?
2. Can we generalize this result to other powers or bases?
3. What happens if the series starts with a different initial term?
4. How can we visualize convergence of alternating series?
5. Why does the sum of a geometric series formula work for infinite terms?

Tip: Always check the ratio $r$ in a geometric series to ensure $|r| < 1$ for convergence.