Evaluate the sum S for the series: sum from n=1
Evaluate the sum $S$ for the series: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2^{n-1}}. $$
Explanation of Method
Rewriting the general term as a power of a single ratio reveals that this is a geometric series $\sum_{n=1}^{\infty} a\,r^{n-1}$ with an alternating, shrinking common ratio. Once the first term $a$ and ratio $r$ are identified, checking $|r|<1$ confirms convergence, and the closed-form sum formula $S = \dfrac{a}{1-r}$ applies directly.
- Identify the series as geometric by rewriting the general term as $a\,r^{n-1}$.
- Check convergence by testing whether $|r| < 1$.
- Apply the geometric series sum formula.
Step 1 · Identify the series type
The general term can be rewritten to reveal its structure:
$$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2^{n-1}} = \sum_{n=1}^{\infty} \left(\frac{-1}{2}\right)^{n-1} = 1 - \frac12 + \frac14 - \frac18 + \cdots $$
This is a geometric series of the form $\sum_{n=1}^{\infty} a\,r^{n-1}$, where:
- The first term is $a = \dfrac{(-1)^{0}}{2^{0}} = 1$.
- The common ratio is $r = -\dfrac12$.
Step 2 · Check for convergence
The series converges if and only if $|r| < 1$:
$$ |r| = \left|-\frac12\right| = \frac12. $$
Since $|r| = \tfrac12 < 1$, the series converges.
Step 3 · Calculate the sum $S$
The sum of a convergent geometric series starting at $n=1$ is:
$$ S = \frac{a}{1-r}. $$
Substituting $a=1$ and $r=-\tfrac12$:
$$ S = \frac{1}{1-\left(-\frac12\right)} = \frac{1}{1+\frac12} = \frac{1}{\frac32} = \frac23. $$
Summary Table
| Quantity | Value |
|---|---|
| First term $a$ | $1$ |
| Common ratio $r$ | $-\dfrac12$ |
| $|r|$ | $\dfrac12 < 1$ (converges) |
| Sum $S$ | $\dfrac23$ |
Additional Notes
- The alternating sign comes entirely from the factor $(-1)^{n-1}$, which is equivalent to raising a negative ratio $r=-\tfrac12$ to successive powers — no separate sign analysis is needed once the series is written as $a\,r^{n-1}$.
- Because $|r|=\tfrac12<1$, the terms shrink geometrically fast, so the partial sums converge quickly toward $\tfrac23$; e.g. after just a few terms the sum is already very close to the limit.
- As a check, note $\tfrac23$ lies between the first two partial sums $1$ and $\tfrac12$, consistent with an alternating series whose partial sums oscillate and converge toward the true sum from both sides.
Result
The sum of the series is: