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

The sum from n=0 to infinity of (0.2)^n is 1.25

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

n = 0 \sum \infty (0.2)^{n}

1.25

Solution steps

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

Apply Series Geometric Test:

1.25

= 1.25

n = 0 \sum \infty \frac{n ^{2}}{9 ^{n}}

n = 0 \sum \infty n \cdot e^{- n}

n = 0 \sum \infty \frac{( - 4 ) ^{n}}{n}

n = 0 \sum \infty \frac{1}{5 ^{n} - 4 ^{n}}

n = 1 \sum \infty (cos (10))^{n}

What is the sum from n=0 to infinity of (0.2)^n ?
The sum from n=0 to infinity of (0.2)^n is 1.25