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Determine the sum of the following series. \sum_{n= 1}^{\infty} \frac{(- 1)^{n- 1}}{5^{n}}
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Step 1:
: Recognize that this is an alternating series with terms in the form of \frac{(- 1)^{n- 1}}{5^{n}}.

Step 2:
: To find the sum of an alternating series, we can use the alternating series test.

The test states that if the limit as n approaches infinity of |a\_n| is zero, then the sum of the series is convergent and is equal to the limit as n approaches infinity of the sum from 1 to n of (- 1)^(n+ 1) * a\_n.

Step 3:
: In this case, |a\_n| is simply \frac{1}{5^{n}}.

We know that the limit as n approaches infinity of $$\frac{1}{5^{n}}$$ is 0, since 5 raised to any power greater than 0 will approach infinity and 1 divided by infinity is 0.

Step 4:
: Now we can apply the formula from the alternating series test:

\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{5^{n}}

Step 5:
: We can simplify this expression by combining the constants (- 1)^(n+ 1) and \frac{1}{5^{n}} into a single fraction:

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{5^{n}} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{5^{n}}

Step 6:
: To find the sum of this series, we can use the formula for the sum of an infinite geometric series, which is \frac{a}{1 -r}, where a is the first term and r is the common ratio.

In this case, a = $$\frac{1}{5}$$ and r = -$$\frac{1}{5}$$.

Step 7:
: Plugging these values into the formula, we get:

\frac{\frac{1}{5}}{1 - (-\frac{1}{5})} = \frac{\frac{1}{5}}{\frac{6}{5}} = \frac{1}{6} \approx 0.1667

Final Answer

The sum of the series \sum_{n= 1}^{\infty} \frac{(- 1)^{n- 1}}{5^{n}} is approximately 0.1667.