What is the sum from n=1 to infinity of 0.4^n ?

The sum from n=1 to infinity of 0.4^n is 0.66666…

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sum from n=1 to infinity of 0.4^n

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

0.66666 \dots

Solution steps

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

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

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

n = 0 \sum \infty 0. 4^{n} = 1.66666 \dots

= 1.66666 \dots - 0. 4^{0}

Simplify

= 0.66666 \dots

d er i v a t i v e f (x) = x^{3} - x^{2}

d er i v a t i v e f (x) = 4 x^{2} + 2

n = 1 \sum \infty \frac{1}{n 7 ^{n}}

t an g e n t 4 x - x^{2}

d er i v a t i v e ln (x + y)

What is the sum from n=1 to infinity of 0.4^n ?
The sum from n=1 to infinity of 0.4^n is 0.66666…