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

\frac{d}{d x} (- \frac{x ^{l n (x)}}{6})

f (x) = x^{2} - 4 x, g (x) = 0

\frac{d}{d x} ((5 x^{7} - 7 x^{5} + 12) (3 x^{6} - 2 x))

x \to 1 lim (x + 3)

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