What is the sum from n=1 to infinity of 4*0.3^n ?

The sum from n=1 to infinity of 4*0.3^n is 1.71428…

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sum from n=1 to infinity of 4*0.3^n

n = 1 \sum \infty 4 \cdot 0. 3^{n}

1.71428 \dots

Solution steps

n = 1 \sum \infty 4 \cdot 0. 3^{n}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty 0. 3^{n}

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

= 4 (n = 0 \sum \infty 0. 3^{n} - 0. 3^{0})

n = 0 \sum \infty 0. 3^{n} = 1.42857 \dots

= 4 (1.42857 \dots - 0. 3^{0})

Simplify

= 1.71428 \dots

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What is the sum from n=1 to infinity of 4*0.3^n ?
The sum from n=1 to infinity of 4*0.3^n is 1.71428…