How do you find the sum of #Sigma(-2/7)^n# from n #[0,oo)#?

Question

Ezra Adams · Accepted Answer

undefined The sum of the series $ \sum_{n=0}^{\infty} \left(-\frac{2}{7}\right)^n $ can be found using the formula for the sum of an infinite geometric series, which is given by:

$$ S = \frac{a}{1 - r} $$

Where:
- $ S $ is the sum of the series,
- $ a $ is the first term of the series, and
- $ r $ is the common ratio of the series.

In this series, $ a = 1 $ (since it starts from $ n = 0 $) and $ r = -\frac{2}{7} $.

Plugging these values into the formula:

$$ S = \frac{1}{1 - \left(-\frac{2}{7}\right)} $$

$$ S = \frac{1}{1 + \frac{2}{7}} $$

$$ S = \frac{1}{\frac{9}{7}} $$

$$ S = \frac{7}{9} $$

Therefore, the sum of the series $ \sum_{n=0}^{\infty} \left(-\frac{2}{7}\right)^n $ is $ \frac{7}{9} $.

Evelyn Agard · Answer

Use the Geometric Series Theorem.

#-2/7# is the common ratio of the Geometric Series in this problem. Since #|-2/7| < 1#, the geometric series converges, and the sum of such a series -- where r is its common ratio -- is given by #sum_0^(oo)(r)^n = 1/(1-r)#

Samantha Adkison · Answer

undefined The sum of the series $ \sum_{n=0}^{\infty} (-2/7)^n $ converges to a finite value if the absolute value of the common ratio $ r $ is less than 1. In this case, the absolute value of $ -2/7 $ is less than 1, so the series converges.

To find the sum, you use the formula for the sum of an infinite geometric series:

$$ S = \frac{a}{1 - r} $$

Where:
- $ a $ is the first term of the series
- $ r $ is the common ratio

In this series:
- $ a = 1 $ (since $ n = 0 $ gives $ (-2/7)^0 = 1 $)
- $ r = -2/7 $

Plug these values into the formula:

$$ S = \frac{1}{1 - (-2/7)} $$

Simplify:

$$ S = \frac{1}{1 + 2/7} $$

$$ S = \frac{1}{7/7 + 2/7} $$

$$ S = \frac{1}{9/7} $$

$$ S = \frac{7}{9} $$

So, the sum of the series $ \sum_{n=0}^{\infty} (-2/7)^n $ is $ \frac{7}{9} $.