To find the sum of the infinite series
[tex]\[
\sum_{k=8}^{\infty} 6\left(\frac{1}{3}\right)^{k-1},
\][/tex]
we need to determine whether this series is a geometric series and then apply the appropriate formula for the sum of an infinite geometric series.
### Step-by-Step Solution
1. Identify the components of the geometric series:
- The general form of a geometric series is [tex]\( ar^{n-1} \)[/tex].
- In our series, the first coefficient [tex]\( a \)[/tex] is 6.
- The common ratio [tex]\( r \)[/tex] is [tex]\( \frac{1}{3} \)[/tex].
- The series starts at [tex]\( k=8 \)[/tex].
2. Determine the first term of the series when [tex]\( k = 8 \)[/tex]:
The first term ([tex]\( a_1 \)[/tex]) when [tex]\( k = 8 \)[/tex] can be found by plugging [tex]\( k = 8 \)[/tex] into the general term:
[tex]\[
6\left(\frac{1}{3}\right)^{8-1} = 6\left(\frac{1}{3}\right)^{7}.
\][/tex]
This is the first term of the series.
3. Calculate the value of the first term:
[tex]\[
6\left(\frac{1}{3}\right)^{7} = 6 \times \left(\frac{1}{2187}\right) = \frac{6}{2187} = 0.0027434842249657054.
\][/tex]
4. Apply the formula for the sum of an infinite geometric series:
The sum [tex]\( S \)[/tex] of an infinite geometric series starting from the given first term ([tex]\( a_1 \)[/tex]) is given by:
[tex]\[
S = \frac{a_1}{1 - r}
\][/tex]
Using our values:
[tex]\[
S = \frac{0.0027434842249657054}{1 - \frac{1}{3}} = \frac{0.0027434842249657054}{\frac{2}{3}} = 0.004115226337448557.
\][/tex]
5. Final Answer:
So, the sum of the infinite series
[tex]\[
\sum_{k=8}^{\infty} 6\left(\frac{1}{3}\right)^{k-1}
\][/tex]
is approximately [tex]\( 0.004 \)[/tex].
Thus, the correct answer to the given question is:
[tex]\[
\boxed{0.004}
\][/tex]
