Questions: What is the sum of the series from n=1 to infinity of 2(1/5)^(n-1)? S=1/5 S=2/5 S=5/3 S=5/2

Solution Steps

To find the sum of the infinite series \(\sum_{n=1}^{\infty} 2\left(\frac{1}{5}\right)^{n-1}\), we recognize that this is a geometric series with the first term \(a = 2\) and common ratio \(r = \frac{1}{5}\). The sum of an infinite geometric series is given by the formula \(S = \frac{a}{1 - r}\).

The given series is \(\sum_{n=1}^{\infty} 2\left(\frac{1}{5}\right)^{n-1}\). This is a geometric series with:

• First term \(a = 2\)
• Common ratio \(r = \frac{1}{5}\)

The sum \(S\) of an infinite geometric series is given by: \[ S = \frac{a}{1 - r} \]

Substitute \(a = 2\) and \(r = \frac{1}{5}\) into the formula: \[ S = \frac{2}{1 - \frac{1}{5}} = \frac{2}{\frac{4}{5}} = 2 \times \frac{5}{4} = 2.5 \]

Final Answer