What is the sum from n=0 to infinity of 2*0.2^n ?

The sum from n=0 to infinity of 2*0.2^n is 2.5

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sum from n=0 to infinity of 2*0.2^n

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

2.5

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

1.25

= 2 \cdot 1.25

Simplify

= 2.5

x \to 0 lim (\frac{5 ^{x + 1} - 5}{x})

\int_{0}^{\infty} 8 sin^{2} (x) d x

\int arctan (d) x d x

\frac{d}{d θ} (\frac{2}{1 - sin ( θ )})

x \to 5 lim (x^{2} + 5 x + 8)

What is the sum from n=0 to infinity of 2*0.2^n ?
The sum from n=0 to infinity of 2*0.2^n is 2.5