To determine whether the given series [tex]\(\sum_{n=1}^{\infty}\left(-\frac{2}{6}\right)^{n-1}\)[/tex] converges or diverges, and to find its sum if it does converge, we can proceed as follows:
1. Identify the type of series:
The given series is an infinite geometric series in the form:
[tex]\[
\sum_{n=1}^{\infty} ar^{n-1}
\][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(r\)[/tex] is the common ratio.
2. Determine the first term ([tex]\(a\)[/tex]) and the common ratio ([tex]\(r\)[/tex]) of the series:
- The first term [tex]\(a\)[/tex] is the value of the series when [tex]\(n = 1\)[/tex]:
[tex]\[
a = \left(-\frac{2}{6}\right)^{1-1} = 1
\][/tex]
- The common ratio [tex]\(r\)[/tex] is given as:
[tex]\[
r = -\frac{2}{6} = -\frac{1}{3}
\][/tex]
3. Check for convergence:
For an infinite geometric series to converge, the absolute value of the common ratio [tex]\(r\)[/tex] must be less than 1:
[tex]\[
|r| < 1
\][/tex]
Here, [tex]\(|r| = \left| -\frac{1}{3} \right| = \frac{1}{3}\)[/tex], which is indeed less than 1.
Therefore, the series converges.
4. Calculate the sum of the infinite geometric series:
If the series converges, the sum [tex]\(S\)[/tex] of the infinite geometric series is given by the formula:
[tex]\[
S = \frac{a}{1 - r}
\][/tex]
Substituting [tex]\(a = 1\)[/tex] and [tex]\(r = -\frac{1}{3}\)[/tex] into the formula, we get:
[tex]\[
S = \frac{1}{1 - \left(-\frac{1}{3}\right)} = \frac{1}{1 + \frac{1}{3}} = \frac{1}{\frac{4}{3}} = \frac{3}{4} = 0.75
\][/tex]
5. Conclusion:
- This series converges.
- The sum of the series is [tex]\(\boxed{0.75}\)[/tex].
Hence, the series [tex]\(\sum_{n=1}^{\infty}\left(-\frac{2}{6}\right)^{n-1}\)[/tex] converges with a sum of [tex]\(0.75\)[/tex].
