What is the sum from n=1 to infinity of 2(0.8)^n ?

The sum from n=1 to infinity of 2(0.8)^n is 8

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sum from n=1 to infinity of 2(0.8)^n

n = 1 \sum \infty 2 (0.8)^{n}

8

Solution steps

n = 1 \sum \infty 2 \cdot 0. 8^{n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty 0. 8^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 2 (n = 0 \sum \infty 0. 8^{n} - 0. 8^{0})

n = 0 \sum \infty 0. 8^{n} = 5

= 2 (5 - 0. 8^{0})

Simplify

= 8

x \to 3 + lim (\frac{2}{( x - 3 ) ^{3}})

1 + \frac{1}{x ^{2}}

\int (2 x - 3 x^{2}) d x

sin ((5 x^{2} + 8)^{8})

\frac{s ^{3} + s + 2}{s ^{4} - 4}

What is the sum from n=1 to infinity of 2(0.8)^n ?
The sum from n=1 to infinity of 2(0.8)^n is 8