What is the sum from n=0 to infinity of (0.2^n)0.8 ?

The sum from n=0 to infinity of (0.2^n)0.8 is 1

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sum from n=0 to infinity of (0.2^n)0.8

n = 0 \sum \infty (0. 2^{n}) 0.8

1

Solution steps

n = 0 \sum \infty 0. 2^{n} \cdot 0.8

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 0.8 \cdot n = 0 \sum \infty 0. 2^{n}

Apply Series Geometric Test:

1.25

= 0.8 \cdot 1.25

Simplify

= 1

k = 7 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

k = 0 \sum \infty (\frac{4}{7})^{k}

x = 1 \sum \infty \frac{x + 2}{x + 1}

n = 1 \sum \infty (\frac{1}{3})^{2 n}

n = 0 \sum \infty \frac{cos ( πn )}{5}

What is the sum from n=0 to infinity of (0.2^n)0.8 ?
The sum from n=0 to infinity of (0.2^n)0.8 is 1